Illlll™ ™
;i!l:l!lill'l!li:!li.ll|l
mil iiiii ii
Dt^ata, ^tm fattt
BOUGHT WITH THE INCOME OF THE
SAGE ENDOWMENT FUND
THE GIFT OF
HENRY W. SAGE
1891
Cornell University Library
olin.anx
3 1924 031 245 1
Cornell University
Library
The original of tliis book is in
tine Cornell University Library.
There are no known copyright restrictions in
the United States on the use of the text.
http://www.archive.org/cletails/cu31924031245156
ELEMENTARY
MATHEMATICAL ANALYSIS
MODERN MATHEMATICAL TEXTS
EDITED BY
Charles S. Slighter
ELEMENTARY MATHEMATICAL ANALYSIS
By CHABLsa S. Slichteb
i97 pages, 5% 7ii,niiiatraled $2,60
MATHEMATICS FOR AGRICULTURAL
STUDENTS
By Henry C. Wolff
311 panes, 5 % 7H, lUustraled . $1.50
CALCULUS
By Herman W. March and Henry C. Wolff
360 pages, 5 I 7H, Illustrated. . $2.00
PROJECTIVE GEOMETRY
, By L. Watland Dowling
316 pages, a ii7H,IllustraUd $2.00
MODERN MATHEMATICAL TEXTS
Edited bt Chaelbs S. Slichtee
ELEMENTARY
MATHEMATICAL ANALYSIS
A TEXT BOOK FOR FIRST
YEAR COLLEGE STUDENTS
BY
CHARLES S. SLICHTER, Sc. D.
PROFESSOR OF APPLIED MATHEMATICS
■DNIVERSITY OP WISCONSIN
Second Edition
Revised and Entirely Reset
McGRAW-HILL BOOK COMPANY, Inc.
. 239 WEST 39TH STREET. NEW YORK
LONDON: HILL PUBLISHING CO., Ltd.
6 & 8 BOUVBKIE ST., E. C-
1918
A-
COPYEIGHT, 1914, 1918, BY THe]
McGeaw-Hill Book Company, Inc.
TMT- MAPtK PUESS roBK FA,
PREFACE TO THE SECOND EDITION
In rewriting the present book, simplification of the material
has been the main end in view. Considerable matter has been
omitted, and numerous worked exercises have been added. The
second chapter is devoted to an introduction to rectangular
coordinates and to the straight line. New sets of exercises and
long lists of miscellaneous and review exercises have been inserted
at appropriate places. Changes in order of material and in
method of treatment have been made freely.
Much greater use has been made of fine print than in the first
edition. Sections of the text which can readily be omitted have a
star attached to the section numbers. Some of these of a sub-
ordinate illustrative character, or primarily intended for reference,
are put in fine print.
The review chapter on elementary algebra has been greatly
enlarged. This material is placed in the last chapter or appendix,
where the considerable amount of very elementary mathematics
will not at once confront and perhaps discourage the well-prepared
student. At the same time enough material is given so that
students with but a single year of high school algebra can be
gotten ready for the course. The elementary material is so
classified that either a few days, or several weeks may be devoted
to the review.
The writer is greatly indebted to many persons for aid in the
revision. Professors March and Wolff of the University of
Wisconsin have contributed much, and Professor Wolff has read
all of the galley proof. I am especially indebted to Professors
Jordan and Lefschetz of the University of Kansas for many
valuable suggestions and to Professor L. C. Plant of Lansing,
Michigan. To all of these my especial thanks are due.
Charles S. Slichter.
FROM THE PREFACE TO THE
FIRST EDITION
This book is not intended to be a text on "Practical Mathe-
matics" in the sense of making use of scientific material and of
fundamental notions not already in the possession of the student,
or in the sense of making the principles of mathematics secondary
to its technique. On the contrary, it has been the aim to give
the ftmdamental truths of elementary analysis as much promi-
nence as seems possible in a working course for freshmen.
The emphasis of the book is intended to be upon the notion of
functionality. Illustrations from science are freely used to make
this concept prominent. The student should learn early in his
course that an important purpose of mathematics is to express and
to interpret the laws of actual phenomena and not primarily to
secure here and there certain computed results. Mathematics
might well be defined as the science that takes the broadest view of
all of the sciences — an epitome of quantitative knowledge. The
introduction of the student to a broad view of mathematics can
hardly begin too early. ,
The ideas explained above are developed in accordance with a
two-fold plan, as follows :
First, the plan is to group the material of elementary analysis
about the consideration of the three fundamental functions:
1. The Power Function y = ax" (n any number) or the law
"as X changes by a fixed multiple, y changes by a fixed multiple also."
2. The Simple Periodic Function y = asin mx, considered as
fundamental to all periodic phenomena.
3. 'The Exponential Function, or the law "as x changes by a fixed
increment, y changes by a fixed multiple."
Second, the plan is to enlarge the elementary functions by the
development of the fundamental transformations applicable to
these and other functions. To avoid the appearance of abstruse-
Viii PREFACE
ness, these transformations are stated with respect to the graphs
of the functions; that is, they are not called transformations, but
"motions" of the loci. The facts are summarized in several
simple "Theorems on Loci," which explain the translation, rota-
tion, shear, and elongation or contraction of the graph of any
function in the xy plane.
Combinations of the fundamental functions as they actually
occur in the expression of elementary natural laws are also dis-
cussed and examples are given of a type that should help to ex-
plain their usefulness.
Emphasis is placed upoji the use of time as variable. This
enriches the treatment of the elementary functions and brings
many of the facts "analytic geometry" into close relation to
their application in science. A chapter on waves is intended to
give the student a broad view of the use of the trigonometric func-
tions and an introduction to the application of analysis to periodic
phenomena.
It is difficult to understand why it is customary to introduce
the trigonometric functions to students seventeen or eighteen years
of age by means of the restricted definitions applicable only to the
right triangle. Actual test shows that such rudimentary methods
are wasteful of time and actually confirm the student in narrow-
ness of view and in lack of scientific imagination. For that reason,
the definitions, theorems and addition formulas of trigonometry
are kept as general as practicable and the formulas are given
general demonstrations.
The possibiUties and responsibiUties of character building in the
department of mathematics are kept constantly in mind. It is
accepted as fundamental that a modern working course in mathe-
matics should emphasize proper habits of work as well as proper
methods of thought; that neatness, system, and orderly habits
have a high value to all students of the sciences, and that a text-
book should help the teacher in every known way to develop these
in the student. '
The present work is a revision and rewriting of a preliminary
form which has been in use for three years at the University of
Wisconsin. During this time the writer has had frequent and
valuable assistance from the instructional force of the department
PREFACE IX
of mathematics in the revision and betterment of the text. Ac-
knowledgments are due especially to Professors Burgess, Dresden,
Hart and Wolfif and to Instructors Fry, Nyberg and Taylor.
Professor Burgess has tested the text in correspondence courses,
and has kindly embraced that opportunity to aid very materially
in the revision. He has been especially successful in shortening
graphical methods and in adapting them to work on squared paper.
Professor Wolff has read all of the final manuscript and made
many suggestions based upon the use of the text in the class room.
Mr. Taylor has read all of the proof and supphed the results to the
exercises.
Professor E. V. Huntington of Harvard University has read the
galley proof and has contributed many important suggestions.
The writer has avoided the introduction of new technical terms,
or terms used in an unusual sense. He has taken the liberty, how-
ever of naming the function ax", the "Power Function of x," as a
short name for this important function seems to be an unfortu-
nate lack — ^a lack, which is apparently confined solely to the
Enghsh language.
Chables S. Slichtbr.
University or Wisconsin
July 25, 1914
Note: The results to the exercises are issued aa a separate pamphlet.
CONTENTS
Faqb
Preface . . v
Intbodtjction .... . . xiii
Mathematical Signs and Symbols. ... . . . xviii
Chaptbr
I. Vakiablbb and Functions op Vabiablbs ... . 1
II. Rectanqtjlak Cookdinatbs and the Straight
Line ... ... .... 23
III. The Powbe Function . 48
MiSCELLANBOITS ExEBCISES. .... 92
IV. The Circle and the Circular Functions .... 97
V. The Ellipse and Htpebbola .152
VI. Single and Simultaneous Equations 174
VII. Permutations, Combinations; the Binomial
Theorem . ... 198
VIII. Progressions . . ... 213
Questions and Exercises for Review, Chapters
I to VIII 225
IX. The Logarithmic and Exponential Functions . 234
X. Tbigonombtbic Equations and the Solution of
Tbiangles 304
XI. Simple Harmonic Motion and Waves . . . 339
XII. Complex Numbebs 357
XIII. Loci 387
XIV. The Conic Sections . . 399
XV. Appendix — A Review of Sbcondabt School
Algbbba 451
Mathematical Tables 474
Index 491
INTRODUCTION
Any course in mathematics requires the frequent use of geo-
metrical constructions, and the carrying out of analytical and
numerical computations. In order that this work may be per-
formed neatly and accurately it is necessary that the student
have a few simple instruments, and a supply of proper material
for doing the work in a systematic and orderly manner. The
indispensible instruments are as follows :
I. Instruments. (1) Two 4:H hexagonal drawing pencils; one
sharpened to a fine point for marking points upon paper or for sketch-
ing free hand; the other sharpened to a chisel point for drawing
straight lines. Some prefer to use a single pencil sharpened at both
ends, one end round pointed, the other end chisel pointed.
(2) A small drawing board' of soft wood — 10X12 inches is large
enough.
(3) A small T-square same length as the drawing board.
(4) A 60° and a 45° transparent triangle. Five-inch triangles
are large enough, although a larger 60° triangle will be found to be
very convenient.
(5) A protractor for laying off angles.
(6) A triangular boxwood scale, decimally divided.
(7) A pair of 6-inoh pencil compasses for drawing circles and
arcs of circles, provided with medium hard lead, sharpened to a
narrow chisel point.
(8) A 10-inch sUde rule is required for Chapter IX, and may be
used earlier at the discretion of the instructor.
II. Materials. All mathematical work should be done on one
side of standard size letter paper, 8^ X 11 inches. This is the
smallest sheet that permits proper arrangement of mathematical
work. A good equipment will include:
(1) A notQ book cover to hold sheets of the above named size and
^Drawing boards of this size with T-square and two wood triangles are marketed
by the Milton Bradly Co., Springfield, Mass.
xiii
XIV INTRODUCTION
a supply of manUa paper "vertical file folders" for use in submit-
ting work for the examination of the instructor.
(2) A number of different forms of squared paper and computa-
tion paper especially prepared for use with this book. These sheets
will be described from time to time as needed in the work. Form
M2 wiU be found convenient for problem work and for general cal-
culation. M2 is a copy of a form used by a number of public utiUty
and industrial corporations. Colleges usually have their own sources
of supply of squared paper, satisfactory for use with this book.
(3) Miscellaneous supphes such as thumb tacks, erasers, sandpaper-
pencil-sharpeners, etc.
in. General Directions. All drawings should be done in pencil,
unless the student has had training in the use of the ruling pen,
in which case he may, if he desires, "ink in" a few of the most
important drawings.
AU mathematical work, such as the solutions of problems and
exercises, and work in computation should be done in ink. The
student should acquire the habit of working problems with pen
and ink. He will find that this habit will materially aid him in
repressing carelessness and indifference and in acquiring neatness
and system.
TO THE STUDENT— SUGGESTIONS ON THE STUDY OF
MATHEMATICS
The following suggestions may assist the student to acquire habits
of work essential to success in the study of mathematics and of the
other exact sciences.
Successful intellectual work depends very largely upon the power
of concentration. Fortxmately this power can be acquired and culti-
vated'. The student should study away from interruption and then
must not permit his work to become interrupted by himseU or by
others. By holding his attention upon his work and by keeping his
mind from wandering to extraneous matters, the student will cultivate
a fundamental habit that will tend to assure his success both in and
out of college.
In a course in mathematics a student (1) studies a textbook and
(2) works exercises and problems. An assigrmient for a given day
may therefore consist of the study of mathematical principles and
theory (such as theorems, definitions, and explanations of processes),
or it may consist of the working out of exercises and problems, or,
as is usually the case, it may consist of the theory and principles of
INTRODUCTION xv
processes, together with an assignment of exercises illustrative of the
theory.
1. THE STUDY OF THE TEXTBOOK
Studying a mathematical textbook involves much more than
the mere reading of the statements of principles and of the explanation
of processes. The student must usually read the assigned paragraphs
several times and must frequently turn back and re-read portions of
the text included in previous lessons. In this manner the various
points in the reasoning or explanations can be thought over, and the
habit of asking self-put questions about the work can be acquired.
First of all, in preparing a lesson, try to find out what 'it is about —
what its purpose is. Try to decide how you yourself would go about
the aocompUshment of the task and, if possible, make an independent
attempt of your own. The more consideration you give to such an
attempt, the greater scientific power you will gain.
In particular:
(A) The student should remember that the words in science have
exact meanings and, of course, these meanings must be known to the
student. In studying mathematics the student should acquire
and use the language oi mathematics. For example, he should not
say ' equation" when he means "expression." Indeed, he should go
farther than this. He should make a conscious effort to use abso-
lutely correct English, not only in written work but in oral work
as well.
(B) While studying the text, work out theorems or illustrative
examples with pen and ink. Do not rely upon a mere reading —
even repeated readings — of a new piece of reasoning or of the explana-
tion of a new process.
(C) Bead over all of the lesson assigned in the text a last time after
working the assigned exercises. The text will probably have a new
meaning after working out the special cases in the exercises. This
habit will give a meaning to the words, "Learn by doing."
(D) Finally, make a mental simimary of each lesson.
(E) Review often.
2. THE WORKING OF EXERCISES
(F) Read each exercise or problem carefully and plan a method
of attack in advance in order to facilitate arrangements of equations
and computations and the drawing of figures.
(G) Look at your result and see if it is a reasonable one.
XVI INTRODUCTION
(H) Check result.
(I) Indicate the results by a distinguishing mark, or summarize
in logical qrder.
(J) The figures and diagrams should have sufficient lettering,
titles, etc., to make them self-explanatory. The units of measure
used should, of course, be clearly indicated.
(K) Do all work neatly the first time and (except drawings)
invariably in ink. Try to have the first draft sufficiently neat in
appearance and arrangement to hand in to your instructor.
(L) After the first draft has been finished, read it over carefully
to see where it may be unproved in method or arrangement and
think about the processes you have used. If small changes only
are needed to effect the desired improvement, make them by drawing
lines through the portions to be changed and by making neat inser-
tions. If considerable changes are necessary, do the work over.
The study and improvement of the work will prove to be of fully
as much importance to you as the doing of the work itself.
(M) See to it that each piece of work or exercise is complete. On
any piece of written work the nature of the problem should be clearly
and briefly stated. The student should learn to think of each piece
of work as a thing that is in itself worth while. Hence each detail
should be attended to before the work is submitted to the instructor.
See that sufficient explanation is given and that the numbers and
magnitudes are adequately named and labelled.
TO THE INSTRUCTOR
The instructor cannot insist too emphatically upon the require-
ment that all mathematical work done by the student — ^whether
preliininary work, numerical scratch work, or any other kind (except
drawings) — shall be carried out with pen and ink upon paper of
suitable size. This should, of course, include all work done at home,
irrespective of whether it is to be submitted to the instructor or not.
The "psychological effect" of this requirement will be found to entrain
much more than the acquirement of mere technique. If properly
insisted upon, orderly and systematic habits of work will lead to
orderly and systematic habits of thought. The final results will be
very gratifying to those who sufficiently persist in this requirement.
At institutions whose requirements for admisstion include more
than one and one-half units of preparatory algebra, nearly all of
Chapters VI, VII, and VIII may be omitted from the course.
An asterisk attached to a section number indicates that the section
INTRODUCTION xvii
may he omitted. These sections will frequently be found useful in
forming the basis of discussion by the instructor.
The usual one and one-half year of secondary school Algebra,
including the solution of quadratic equations and a knowledge of
fractional and negative exponents, is required for the work of this
course. In the appendix (Chapter XV) vrill be found material for a
brief review of factoring, qitadratics, and exponents, upon which a week
or ten days should be spent before beginning the regular work in this
text.
This review chapter is placed last because the amount of material
in it is greater than need be taken in all cases and also because college
students do not like to be confronted on the &st page of a scientific
text-book with elementary work of high school grades.
GREEK ALPHABET
Capitals
Lower
case
Names
Capitals
Lower
case
Names
A
a
Alpha
N
V
Nu
B
P
Beta
S
i
Xi
r
y
Gamma
0
o
Omicron
A
s
Delta
II
IT
Pi
E
e
Epsilon
p
9
Rho
Z
f
Zeta
s
a
Sigma
, H
V
Eta
T
T
Tau
e
e
Theta
T
V
Upsilon
I
L
Iota
*
<t>
Phi
K
K
Kappa
X
X
Chi
A
X
Lambda
*
i'
Psi
M
/<
Mu
Q
U>
Omega
MATHEMATICAL SIGNS AND SYMBOLS
read
=
read
5^
read
=
read
=F
read
>
read
<
read
1^
read
(a,b)
read
|n
read
n!
read
limit r,, ,'
read
a; = oo
read
\a\
read
log„a;
read
Iga;
read
In a;
read
and so on.
is identical vnth.
is not equal to.
approaches.
is approximately equal to.
is greater than.
is less than.
is greater than or equal to.
point whose coordinates are a and b.
factorial n.
factorial w or n admiration.
limit of fix) as x approaches a.
X becomes infinite,
absolute value of a.
logarithm of x to the base a.
common logarithm of x.
natural logarithm of x.
Sw„
read summation from n = 1 io n = r of u„
ELEMENTARY
MATHEMATICAL ANALYSIS
CHAPTER I
VARIABLES AND FUNCTIONS OF VARIABLES
1. Scales. Select a series of points along any curve and mark
the points of division with the numbers of any sequence.' The
result of such a construction is called a scale. Thus in Fig. 1
the points along the curve OA have been selected and marked in
order with the numbers of the sequence:
Oj 4j 2> 1> 25, 3, 5, 7, 8
A non-uniform scale.
Thus primitive man might have made notches along a twig
and then made use of it in making certain measurements of
interest to him. If such a scale were to become generally used by
others, it would be desirable to make many copies of the original
scale. It would, therefore, be necessary to use a twig whose shape
could be readily duplicated; such, for example, as a straight
stick; and it would also be necessary to attach the same symbols
invariably to the same divisions.
Certain advantages are gained (often at the expense of others,
however) if the distances between consecutive points of division
are kept the same; that is, when the intervals are laid off by repe-
tition of the same selected distance. When this is done, the scale
is called a uniform scale. Primitive man might have selected for
1 A sequence of numbers here means a set of numbers arranged in order of
magnitude.
1
2 ELEMENTARY MATHEMATICAL ANALYSIS [§1
such uniform distance the length of his foot, or sandal, the breadth
of his hand, the distance from elbow to the end of the middle
finger (the cubit), the length of a step in pacing (the yard), the
amount he can stretch with both arms extended (the fathom),
etc.
Fig. 2. — An ammeter scale.
We are familiar with many scales, such as those seen on a
yardstick, the dial of a clock, a thermometer, a sun-dial, a steam-
gage, an ammeter or voltmeter, the arm of a store-keeper's
scales, etc. The scales on a clock, a yardstick, or a steel tape are
uniform. Those on a sun-dial,
on some ammeters or on a good
thermometer, are not uniform.
One of the most important
advantages of a uniform scale
is the fact that the place of
beginning, or zero, maybe taken
at any one of the points of divi-
sion. This is not true of a non-
uniform scale. If a sun-dial is not properly oriented, it is useless.
If the needle of an ammeter be bent the instrument cannot be used.
It is always necessary in using such an instrument to know that
the zero is correct. If, however, a yarfistick or a steel tape be
broken, it may stUl be used for measuring lengths. The student
P.M
A.M.
Sun-dial scale.
§2] VARIABLES AND FUNCTIONS OF VARIABLES 3
may think of many other advantages gajned in using a uniform
scale.
2. Formal Definition of a Scale. If points be selected in order
along any curve corresponding, one to one, to the numbers of
any sequence, the curve, with its divisions, is called a scale.
The notion of one to one correspondence, included in this
definition, is frequently used in mathematics.
Ii II I h I II h I I I I 111 il I M I I II II h I I I h I I I h I I I I II I I I
0 1.2 3 1 5
Fig. 4. — A uniform arithmetical scale.
In mathematics we frequently speak of the arithmetical scale
and of the algebraic scale. The arithmetical scale corresponds
to the numbers of the sequence
0, 1, 2, 3, 4, 5, . .
and such intermediate numbers as may be desired. It is usually
represented by a uniform scale as in Fig. 4. The algebraic
scale corresponds to the numbers of the sequence
. . . -6, -5, -4, -3,-2,-1,0,+!, +2, +3, +4, +5, . . .
and such intermediate numbers as may be desired. It is usually
1 I I I I I I I 1 I I I I I I I 11 I I I M I I I I I I I I I I I I I I I I I I I II I I II I I 1
-B -4 -S -2 -1 0 +1 +2 +3 +4 +5
Fig. 5. — A uniform algebraic scale.
represented by a uniform scale as in Fig. 5. The arithmetical
scale begins at 0 and extends indefinitely in one direction. The
algebraic scale has no point of beginning; the zero is placed at any
desired point and the positive and negative numbers are then
attached to the divisions to the right and the left, respectively,
of the zero so selected. The algebraic scale extends indefinitely
in both directions.
Exercises
1. On a uniform algebraic scale, how far is the point marked 5 from
the point marked 7? How far is the point marked 6 from the point
marked 10.5? How far is th'e point marked —10.8 from the point
marked 13.6?
4 ELEMENTARY MATHEMATICAL ANALYSIS [§3
2. Show that the distance between two points selected anywhere on
the uniform algebraic scale is always found by subtraction.
3. What points of the uniform algebraic scale are distant 5 from the
point 3 of that scale? What point of the uniform arithmetical scale
is distant 5 from the point 3 of that scale?
4. -If two algebraic scales intersect at right angles, the common
point being the zero of both scales, explain how to find the distance
from any point of one scale to any point of the other scale.
3. Two Uniform Scales in Juxtaposition or Double Scales.
The relation between two magnitudes or quantities, or between
two numbers, may be shown conveniently by placing two scales
side by side. Thus the relation between the number of centi-
meters and the number of inches in any length may be shown
by placing a centimeter scale and a foot-rule side by side with
their zeros coinciding as in Fig. 6. From this figure it is seen
that 1 inch corresponds to 2.6 centimeters; 3.3 inches correspond
to 8.44 centimeters, 4.6 inches corresponds to 11.76 centimeters;
that 5 centimeters correspond to 1.97 inches, 8.5 centimeters
corresponds to 3.32 inches, etc.
A thermometer is frequently seen bearing both Fahrenheit and
the centigrade scales. See Fig. 7. It is obvious that the double
scale of such a thermometer may be used (within the limits of its
range) for converting any temperature reading Fahrenheit into
the corresponding centigrade equivalent or vice versa. From
Fig. 7 it is seen that 72°F. corresponds to 22.2°C., 212°F. to
100°C., 32°F. to zero degrees centigrade; that 21°C. corresponds
to 69.8°F., 72''C. to 161.6°F.
The construction of scales of the kind considered above
may be made to depend upon the following problem in elementary
geometry : To divide a given line into a given number of equal parts.
Illustration. In Fig. 9 is given a double scale OA-OB showing the
correspondence between speed expressed in miles per hour and speed
expressed in feet per second. The student will reproduce neatly and
accurately the drawing, on a larger scale, in accordance with the
directions given below,
A mUe contains 5280 feet, an hour contains 3600 seconds. Hence,
one mile per hour equals |f-§-J or -f-| feet per second. Therefore,
if two uniform scales. Fig. 9, one rejJresenting speed expressed in
feet per second, and the other representing speed expressed in miles
§3] VARIABLES AND FUNCTIONS OF VARIABLES
-r-S
_ — CO o
-pas
o «
I
a «3
6
M
o o c3
&0
I
- ag
o V
■ g g
=• »•§
•a ^ I
S "» _
O
I- _
5? fe
o
o
00
f=(
6 ELEMENTARY MATHEMATICAL ANALYSIS [§3
20-
B-
18-
11-
i,;i2-
1^
6-
/o.
Fia. 9. — Method of constraction of double scale showing relation
between "miles per hour" and "feet per second."
§3] VARIABLES AND FUNCTIONS OF VARIABLES 7
per hour, are constructed with their zeros coinciding and with the
point marked 22 of the first coinciding with the point marked 15 of the
second, the double scale may be used for converting speed expressed
as miles per hour into speed expressed as feet per second, or vice versa.
Lay off with a scale a line OA 11 inches long. Divide this line
into 22 equal parts, and subdivide each division into 6 equal parts.
Mark these divisions and subdivisions as indicated in Fig. 9. Draw
the line OK, making the angle KOB about 30°. With a pair of bow
dividers or with a scale lay off on OK 15 equal divisions, about f of
an inch each. Let the last point of division be marked C. OC is
then divided into 15 equal parts. Draw CA. With a pair of triangles
draw lines through the points Ci, Ci, Cz, . Cn parallel to CA,
intersecting the line OA in the points marked 15, 14, 13, 1,
respectively. Why is OB a uniform scale divided into 15 equal parts?
Mark the scales OA and OB in red ink with a new set of numbers
so that the double scale may also be used for converting speeds if
the readings fall between 15 and 30 feet per second instead of between
0 and 15.
From the double scale just constructed, find the speeds expressed
as miles per hour corresponding to speeds of 2, 4, 5, 11, 14, 20, and 25
feet per second.
The lengths selected to represent the various units in any dia-
gram are, of course, arbitrary. As, however, the student is
expected to prepare the various constructions and diagrams
required for the exercises in this book on paper of standard
letter size (that is, 8 J by 11 inches), the units selected should
be such as to permit a convenient and practical construction
upon sheets of that size.
Exercises
The student is expected to carry out the actual construction of only
one of the double scales described in the following exercises.
1. Draw a double scale showing the relation between pressure
expressed as inches of mercury and as feet of water, knowing that
the density of mercury is 13.6 times that of water.
These are two of the common ways of expressing pressure. Water
pressure at water power plants, and often for city water service, is
expressed in terms of head in feet. Barometric pressure, and the
vacuum in the suction pipe of a pump and in the exhaust of a con-
densing steam engine are expressed in inches of mercury. The
approximate relations between these units, i.e., 1 atmosphere = 30
8 ELEMENTARY MATHEMATICAL ANALYSIS [§3
inches of mercury = 32 feet of water' = 15 pounds per square inch,
are known to every student of elementary physics. To obtain, in
terms of feet of water, the pressure equivalent of a; feet of mercury,
multiply X by 13.6, the specific gravity of mercury. This product
divided by 12, or 1.13a;, gives the number of feet of water corre-
sponding to X inches of mercury.
If we let the scale of inches of mercurj' range from 0 to 10, then the
scale of feet of water must range from 0 to 11.3. Hence draw a line
OA 10 inches long divided into inches and tenths to represent inches
of mercury. Draw any line OC through 0 and lay off on it 11.3 uni-
form intervals (uich intervals will be satisfactory). Connect the
end division on OA with the end division on OC by a line AC. Then
from 1, 2, 3, inches on OC draw parallels to AC, thus forming
adjacent to OA the scale of equivalent feet of water. Each of these
intervals can then be subdivided into 10 equal parts corresponding
to tenths of feet of water.
2. Draw a double scale showing pressure expressed as feet, of water,
and as pounds per square inch, knowing that one cubic foot of water
weighs 62.5 pounds.
The weight of one cubic foot of water, 62.5 pounds, divided by 144,
the number of square inches on one face of a cubic foot, gives 0.434
pounds per square inch as the equivalent of one foot of water
pressure.
One pound per square inch is equivalent, therefore, to 1/0.434 or
2.30 feet of water pressure. If we let the scale of pounds range from
0 to 10, we may select 1 inch as the equivalent of 1 pound per
square inch, and divide the scale OA into inches and tenths to repre-
sent this magnitude. Draw OC through 0, and lay off 23 uniform
intervals on OC, 1/2 inch being a convenient length for each of these
parts. Connect the end division of OC with A and through all
points of division of OC draw lines parallel to CA. The range may
be extended to any amount desired by annexing ciphers to the
numbers attached to the two scales.
Extending the range by annexing ciphers to the attached numbers
is obviously practicable so long as the various intervals or units are
decimally subdivided. The- method is impracticable for scales that
are not decimally subdivided, such as shilUngs and pence, degrees and
minutes, feet and inches, etc.
3. Draw a double scale showing the relations between cubic feet,
and gallons. One gallon equals 231 cubic inches, but use the
approximate relation, 1 cubic foot equals 71 gallons. Divide the
§4] VABIABLES AND FUNCTIONS OF VARIABLES 9
scale of cubic feet into tenths, the scale of gallons into fourths to
correspond to quarts.
It is obvious that it is always necessary first to select the range
of the various scales, but it is quite as well in this case to show the
equivalents for 1 cubic foot only, as numbers on the various scales
can be multiplied by 10, 100, or 1000, etc., to show the equivalents for
larger amounts.
Select 10 inches = 1 cubic foot for the scale (OA) of cubic feet.
Draw the line OC. On OC lay off 71 equal parts (say, 7^ inches).
Connect the end division with A and draw the parallel lines exactly
as with previous examples. The intervals of the scale of gallons can
then be subdivided into the four equal parts to show quarts.
4. Draw a double scale showing the relation between cubic feet
and liters. One cubic foot equals 28| liters.
5. If a double scale be drawn on a deformable body, as, for example,
on a rubber band, would the double scale still represent true relations
when the rubber band is stretched? What if the stretching were
not uniform?
6. From Fig. 9, find the number of miles per hour corresponding
to 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15 feet per second.
Place the results in tabular form, i.e., in the first of two adjacent
vertical columns place the numbers 0, 1, 2, . . .15; opposite these
numbers place in the second vertical column the corresponding
numbers representing speed as mUes per hour. Give the first vertical
column the heading "Speed-ft./sec," and the second column the
heading "Speed-mi./hr." Aa the speed changes from 1 foot per
second to 2 feet per second, the speed changes by what amount in
miles per hour? As the speed changes from 3 feet per second to 4
feet per second, the speed in miles per hour changes by what amount?
The change in speed as miles per hour is how many times the change in
the speed as feet per second?
4. A Non-uniform Scale in Juxtaposition with a Uniform scale.
Each scale of the double scales constructed in the preceding
section were uniform scales. The construction of a double scale
of this kind was possible because the change in the number of units
of one magnitude represented was directly proportional to the
corresponding change in the number of units of the other
magnitude. It will, however, be sometimes desirable to construct
double scales in which this proportionality does not exist. For
example, if a double scale were to be constructed showing the
10 ELEMENTARY MATHEMATICAL ANALYSIS [§5
relation between the radius and the area of a circle, the preceding
construction could not be used, since the change in area is pro-
portional to the change in the square of the radius and not
to the change in the radius. In this case both scales cannot
be uniform. Figure 10 is a double scale representing the relation
between the radius and the area of a circle. The area is repre-
sented by the points on the uniform scale, the radius by the points
on the non-uniform scale. The relation is A = irr^ where r is
the radius in feet and A is the area in square feet.
KadluB ol circle
4 5 6
iMiliMil[iiil||MliiiilMiJ|iMl|iii[iiiiliii-rlm^ilyi|li|l|ll||-|l|iyl^il|li^^
Illlllllll llllllllllu
Area of Circle
Fig. 10. — Double scale showing the relation between the area of a
circle and its radius.
5. Functions. The relation between two magnitudes expressed
graphically by two scales drawn in juxtaposition, as above, may
sometimes be expressed by means of an equation. Thus, F,
the number representing the degrees Fahrenheit in a temperature
reading, and C, the number representing the degrees centigrade
of the same temperature, are connected by the equation
F = iC + 32. (1)
Again y, the number representing speed measured as miles per
hour, and x, the number representing speed measured as feet
per second, are connected by the equation
y = iU. (2)
Again u, the number representing pressure measured as feet of
water, and v, the number representing the same pressure measured
as pounds per square inch, are connected by the equation
u = U^j^- (3)
Again A, the number representing area of a circle measured as
square feet, and r, representing the radius measured as feet, are
connected by the relation A = irr^. (4)
Note. The letters F, C, x, y, u, v, in the above equations stand
for numbers; -to make this emphatic we sometimes speak of them as
§5] VARIABLES AND FUNCTIONS OF VARIABLES 11
pure or abstract numbers. These numbers are thought of as arising
from the measurement of a magnitude or quantity by the appUoation
of a suitable unit of measure. Thus from the magnitude or quantity
of water, 12 gallons, arises, by use of the unit of measure the gallon,
the abstract number 12.
Algebraic equations express the relation between numbers, and it is
understood that the letters used in algebra stand for numbers and
not for quantities or magnitudes.
Quantity or Magnitude is an answer to the question: "How
much?" Number is an answer to the question: "How many?"
An interesting relation is given by the scales in Fig. 8. This
diagram shows the fee charged for money orders of various
amounts. The amount of the order may first be found on the
upper scale and then the amount of the fee may be read from the
lower scale. The relation here exhibited is quite different from
those previously given. For example, note that as the amount of
the order changes from $50.01 to $60 the fee does not change, but
remains fixed at 20 cents. Then as the amount of the order
changes from $60.00 to $60.01, the fee changes abruptly from
20 cents to 25 cents. For an order of any amount there is a cor-
responding fee, but for each fee there corresponds not an order of
a single value, but orders of a considerable range in value. This is
quite different from the cases presented in Fig. 7. There for each
reading Fahrenheit corresponds a certain reading centigrade,
or vice versa, and for any change, however small, in one of the
temperature readings a change, also small, takes place in the
other reading. For this reason the latter number is said to be
continuous.
The relation between the temperature scales has been expressed
by an algebraic equation. The relation between the value of a
money order and the corresponding fee cannot be expressed by a
similar equation. If we had given only a short piece of the centi-
grade-Fahrenheit double scale, we could, nevertheless, produce it
indefinitely in both directions, and hence find the corresponding
readings for all desired temperatures. But by knowing the fees
for a certain range of money orders we cannot determine the
fees for other amounts. In both of these cases, however, we
express the fact of dependence of one number upon another
12 ELEMENTARY MATHEMATICAL ANALYSIS [§6
number by sajnng that the first number is a function of the second
number.
6. Definition. Any number, u, is said to be a function of
another number, t, if, when the value of t is given, the value of u
is determined. The number t is called the argument of the
function u.
Illustrations. The length of a rod is a function of its tempera-
ture. The area of a square is a function of the length of a side.
The area of a circle is a function of its radius. The square root
of a number is a function of the number. The strength of an
iron rod is a function of its diameter. The pressure in the ocean
is a function of the depth below the surface. The price of a
railroad ticket is a function of the distance to be travelled.
The temperature Fahrenheit is a function of the temperature
centigrade.
It is obvious that any mathematical expression is, by the above
definition, a function of the letter or letters that occur in it.
Thus, in the equations
u = t^ + it + l
_ t - 1
" ~ 2( + 2
u = Vt + 4: + P -\
u is in each case a function of t.
Goods sent by freight are classified into first, second, third,
fourth, and fifth classes. The amount of freight on a package is
a function of its class. It is also a function of its weight. It is
also a function of the distance carried. Only the second of these
functional relations just named can readily be expressed by an
algebraic equation. It is possible, however, to express all three
graphically by means of parallel scales. The definition of the
function is given (for any particular railroad) by the complete
freight tariff book of the railroad.
The fee charged for a money order is a function of the amount of
the order. The functional relation has been expressed graphically
in Fig. 8. Note that for orders of certain amounts, namely,
$2i $5, $10, $20, $30, $40, $50, $60, $75, the function is not de-
fined. The graph alone cannot define the function at these values,
§6] VARIABLES AND FUNCTIONS OF VARIABLES 13
as one cannot know whether the higher, the lower, or an inter-
mediate fee should be demanded. One can, however, define
the function for these values by the supplementary statement
(for example): "For the critical amounts, always charge the
higher fee." As a matter of fact, however, the lower fee is always
charged.
A function having sudden jumps like the one just considered,
is said to be discontinuous.
Illustration 1. One side of a rectangle is 2 centimeters. The other
side is (x + 2) centimeters. Express the area A of the rectangle as
a function of x.
The area is the product of the breadth by the length of the rectangle.
Hence
A =2(x + 2) =2x + 4, (l;
which is the function of x sought.
Illustration 2. The hypotenuse of a right triangle is 10 inches.
One side is x inches. Express A, the area of the triangle, as a function
of x.
Since the hypotenuse squared equals the sum of the squares of the
two legs, we may write
102 = a;2 -f yi^ (1)
where y stands for the length in inches of the second leg of the triangle.
But we know that
A = kxy. (2)
From (1)
y = Vl02 - xS ' (3)
Substituting in (2), we have
A = kxy/im -x\ (4)
which is the function of x desired. ,
Illustration 3. Express the amount A of $1 at simple interest at
6 per cent, for n years as a function of n.
The interest on $1 for n years equals Sy^Tfre. Hence the amount
(which is the principal plus the interest) is' expressed by
,'' A = 1 + TBTO.
Exercises
In the following exercises the function described can be represented
by a mathematical expression. The problem is to set up the expres-
sion in each case.
14 ELEMENTARY MATHEMATICAL ANALYSIS [§7
1. One side of a rectangle is 10 feet. Express the area il as a func-
tion of the other side x.
2. One leg of a right triangle is 15 feet. Express the area .A as a
function of the other leg x.
3. The base of a triangle is 12 feet. Express the area as a function
of the altitude I.
4. Express the circumference of a circle as a function (1) of its
radius r; (2) of its diameter d.
6. Express the diagonal doia, square as a function of one side x,
6. One leg of a right triangle is 10. Express the hypotenuse h as
a function of the other leg x.
7. A Ship B sails on a course AB perpendicular to OA. If OA = 30
mUes, express the distance of the ship from 0 as a function of AB.
8. A circle has a radius 10 units. Express the length of a chord
as a function of its distance from the center.
9. An isosceles triangle has two sides each equal to 15 centimeters,
and the third side equal to x centimeters. Express the area of the
triangle as a function of x.
10. A right cone is inscribed in a sphere of radius 12 inches. Ex-
press the volume of the cone as a function of its altitude I.
Hint: The distance from the center of the sphere to the base of
the cone is (t— 12), if I >12. The radius of the base of the cone is
Vl2'-(.l-12)' or V24J-Z2. What if 2 < 12?
11. A right cone is inscribed in a sphere of radius a. Express the
volume of the cone as a function of its altitude I.
12. One dollar is at compound interest for 20 years at r per cent.
Express the amount A as a function of r.
7. Functional Notation. The following notation is used to ex-
press that one number is a function of another; thus, if u is a
function of t we write
Likewise
y = /W
means that y is a function of x. Other symbols commonly used to
express functions of x are :
Hx), Xix), f'(x), F(x), etc.
These may be read the "(^-function of x," the "Z-function of x,"
etc., or more briefly, "the <l> of x," "the X of x," etc.
Expressing the fact that temperature reading Fahrenheit (.F) is
§8] VARIABLES AND FUNCTIONS OF VARIABLES 15
a function of temperature reading centigrade (C), we may write:
F=f(.C).
This is made specific by writing
F = iC + 32.
Likewise the fact that the charge for freight is a function of class,
weight, and distance, may be written
r = fie, w, d).
To make this functional symbol explicit, might require that we be
furnished with the complete schedule as printed in the freight tariff
book of the railroad. The dependence of the tariff upon class and
weight can usually be readily expressed, but the dependence upon
distance often contains arbitrary elements that cause it to vary
irregularly, even on different branches of the same railroad. A
complete specification of the functional symbol / would be con-
sidered given in this case when the tariff book of the railroad was in
our hands.
8. Variables and Constants. In elementary algebra, a letter is
always used to stand for a number that preserves the same value
in the same problem or discussion. Such numbers are called
constants. In the discussion above we have used letters to stand
for numbers that are assumed not to preserve the same value but
to change in value; such numbers (and the quantities or
magnitudes which they measure) are called variables.
If r stands for the distance of the center of mass of the earth from
the center of mass of the sun, r is a variable. In the equation s =
igt' (the law of falling bodies), if i be the elapsed time, s the distance
traversed from rest by the falling body, and g the acceleration due to
gravity, then s and t are variables and g is the constant 32.2 feet per
second per second.
The following are constants: Ratio of the diameter to the circum-
ference in any circle; the electrical resistance of pure copper at 60° F. ;
the combining weight of oxygen; the density of pure iron; the velocity
of light in empty space.
The following are variables: the pressure of steam in the cyhnder
of an engine; the price of wheat; the electromotive force in an alter-
nating current; the elevation of groundwater at a given place; the
discharge of a river at a given station. When any of these magnitudes
16
ELEMENTARY MATHEMATICAL ANALYSIS
[§9
are assumed to be measured, the numbers resulting are also variables.
The volume of the mercury in a common thermometer is a variable;
the mass of mercury in the thermometer is a constant.
9.* Graphical Computation. The ordinary operations of arith-
metic, such as multiplication, division, involution and evolution,
can be performed graphically as explained below. The graphical
construction of products and quotients is useful in many problems
of science. The fundamental theorem in all graphical computa-
tion is : The homologous sides of similar triangles are in proportion.
Its application is very simple, as wiU appear from the following
work.
Fboblem 1 : To compute graphically the product of two numbers.
Let the two numbers whose product is required be a and b. On
any line lay off the unit y jj
of measurement, 01, Fig. i"
11. On the same line,
and, of course, to the same
,
,f
/
/
/
/
1
1
U)
Ai
r=c
A-
B
/
/
(B)
\l
=
AC
OA
/
/
B
/
\^
O 1
Fig. 11. — Graphical
multiplication by proper-
ties of similar triangles.
5 6 7 8 9 10
Fig. 12.— Method of graphical mul-
tiplication and division carried out on
squared paper. The figure shows 1 . 9
X4.4 = 8.4.
scale, lay off OA equal to one of the factors a. On any other
line passing through 1 lay off a Une IB equal to the other factor
6. Join OB and produce it to meet AC drawn parallel to IB.
Then AC is the required product. For, from similar triangles.
or
AC:\B = 0A: 01,
AC = OA X IB.
(1)
(A)
§9] VARIABLES AND FUNCTIONS OF VARIABLES 17
AC is to be measured with the same scale used in laying off
01, OA, and IB. The number of unit's in AC is then the product
of a by 6.
It is obvious that the angle OiB may be of any magnitude.
Hence it may conveniently be taken a right angle, in which case the
work may readily be carried out on ordinary squared paper.
Many prefer, however, to do the work on plain paper, la3dng off the
required distances by means of a boxwood triangular scale. If
squared paper is preferred draw the two lines OX and OF at
right angles and the unit line If/, as shown in Fig. 12.'!
In the exercises that follow the dimensions are given in inches.
If the centimeter scale or squared paper Form M\ be used, use
2 cer\timeters everywhere in place of 1 inch.
Exercises
1. Find graphically the product of 1.63 by 2.78.
Hird: Choose 2 inches to represent one unit. Draw a horizontal
line OA 5.56 inches long. Lay off the distance 01 2 inches in
length. Draw IB perpendicular or nearly perpendicular to OA and
lay off IB equal to 3.26 inches in length. Draw OB. Draw AC
parallel to IB. Measure AC. One-half of the length of AC in inches
win be the desired product. It will be noticed that the smaller factor
is laid off on IB.
2. Find graphically the product of 3.15 by 6.27. Let 1 inch
represent one unit.
3. Fmd graphically the product of 36.7 by 5.82.
Hivi: Find the product of 3.67 by 5.82 and then move the decimal
point one place to the right.
4. Find graphically the product of 936 by 3.17.
HiTii: VmA the product of 0.936 by 3.17 and move the decimal
point three places to the right in the result obtained. Let 2 inches
represent one unit.
5. Fiud graphically the product of 9.36 by 7.23.
Hint: Ymd the product of 0.936 by 0.723 and move the decimal
point three places to the right in the result obtained. Let 5 inches
represent one unit.
Problem 2 : To compute graphically the quotient of two numbers
a and b. Formula (A) above can be written
18 ELEMENTARY MATHEMATICAL ANALYSIS [§9
From this it is seen that the quotient of two numbers a and 6 can
readily be computed graphically by use of Figs. 11 or 12.
Exercises
1. Compute graphically the quotient of 1.33 divided by 1.72.
Hint: Let 5 inches represent one unit. Lay off OA equal to 8.6
inches. Draw AC perpendicular or nearly perpendicular to OA.
Lay off AC equal to 6.65 inches. Draw OC. Draw IB parallel to
AC. One-fifth the number of inches in the length of IB is the required
quotient.
2. Compute graphically the quotient of 7.32 divided by 1.26.
Hint: Find the quotient of 0.732 by 1.26, using 5 inches to represent
one unit.
3. Compute graphically 137 divided by 732.
Hint: Calculate 1.37 divided by 0.732 and move the decimal point
one place to the left in the result obtained. Use 5 inches to represent
one unit.
Pboblem 3 : To compute graphically the square of any number N.
This is a special case of Problem 1, when the two factors are
equal.
Exercises
1. Find graphically the square of (o) 5; (6) 3; (c) 2,
Hint: In finding the square of 5, first find square of^O.S. Let 10
inches represent one unit.
2. Find graphically the square of 93.6.
Hint: Find the square of 0.936.
3. Find graphically the square of 0.0672.
Hint: Find the square of 0.672.
4. Find graphically the square of 112.
Hint: Find the square of 1.12.
Phoblbm 4 : To compute graphically the reciprocal of any numherN.
This is a special case of Problem 2, when the dividend is 1 and
the divisor is N.
Exercises
Find graphically the reciprocals of the following: (o) 2; (b) 3.5;
(c) 12.3; (d) 0.817. ,
Peoblem 5: To compute graphically the square root of any
VARIABLES AND FUNCTIONS OF VARIABLES
19
number N. On OX, Fig. 13, lay off 01 = 1 and lA = N. Upon
OA as diameter describe a semicircle OCA. At 1 erect a per-
pendicular, IC, to OA. Then IC is the square root of lA.
Another construction is to place a celluloid triangle in the
position shown in Fig. 13, so that the two edges pass through
0 and A and the vertex of the right angle Ues on the line 1 U.
Fig. 13 shows the construction for \/7.
in'
—
u
q
g
■^
-^
^
s
//'
■^-
\
B
^~
A
X
"<
, i
/
.
> (
"^
w
b
/
r
"■-
,
'-
-._
/
1
->.
"
~-
/
f
/
—
—
—
Fig. 13. — Graphical method of the extraction of square roots. The
figure shows Vt = 2 . Q5.
Exercises
Find graphically the square roots of the following: (o) 2; (b) 3;
(c) 5; (d) 10; (e) 932.
Hivi: In part (e) find the square root of 9.32 and move the decimal
point one place to the right in the result obtained.
Problem 6: To compute graphically the integral powers of
any number N. This problem is solved by the successive applica-
tion of Problem 1 to construct N'^, N^, N*, etc., and of Problem 2
to construct iV"', N~^, N'^, etc. This construction is shown for
the powers of 1.5 in Fig. 14.
Exercises
1. Compute graphicaUy (a) (1.2)^; (6) (0.85)»; (c) (1.72)-2.
Hint: Let 5 inches represent one unit.
20
ELEMENTARY MATHEMATICAL ANALYSIS
2. Show that (1.05)^^ = 2.08, so that money at 5 percent compoimd
interest more than doubles itself in fifteen years.
Note: The work is less if (1.05)* is firstfound and then this result
cubed.
3. From the following outline the student is to produce a complete
method, including proof, of constructing successive powers of any
number.
R
4
/
/
/
/
/
/
//
7
3
1
/
/
A
2
/
//
/
1
//
//
/
1
4
.1
1
^
-3
-4
0 1 N 2 3 i
Fig. 14. — Graphical computation of (1.5)" for n = —4,
0, 1, 2, 3, 4, 6.
-3, -2, -1,
Let OA (Fig. 15) be a radius of a circle whose center is 0. Let
OB be any other radius making an acute angle with OA. From B
drop a perpendicular upon OA, meeting the latter at Ai. From Ai
drop a perpendicular upon OB meeting OB at Ai. From .Aj drop a
perpendicular upon OA meeting OA at A3, and so on indefinitely.
Then, if OA be unity, OAi is less than unity, and OAi, OAs, OAt
. . . are, respectively, the square, cube, fourth power, etc.. of OAi.
§10] VARIABLES AND FUNCTIONS Of VARIABLES 21
Instead of the above construction, erect a perpendicular to OB
meeting OA produced at ai. At Oi erect a perpendicular meeting OB
produced at 02, and so on indefinitely. Then if OA be unity, ai is
greater than unity and az, 03, 04, . are, respectively, the square.
Fig. 15.= — Graphical computation of powers of a number.
cube, etc., of oi. As an exercise, construct powers of 4/5 and of 2.5.
4. Show that the successive "treads and risers" of the steps of the
"stairways" of Fig. 16a and 166 are proportional to the powers of r.
The figures are from Milaukovitch, Zeitschrift fiir Math, und Nat.
Unterricht, Vol. 40, p. 329.
Fig. 16. — Computation of ar, or', ar', . . . for r < 1 and for r > 1.
10.* Double Scales for Several Simple Algebraic Functions. We
may make use of the graphical method of computation explained
above to construct graphically double scales representing simple
22
ELEMENTARY MATHEMATICAL ANALYSIS [§10
algebraic relations. For example, we may construct a double
scale for determining the square of any desired number. Call
OA (see Fig. 17) the scale on which we desire to read the number;
call OB the scale on which we read the square. Let us agree to
lay off OA as a uniform scale, using 01 as the unit of measure.
Since we desire to read opposite 0, 1, 2, 3, . . . of the uniform
scale, the squares of these numbers, the lengths along the scale
OB must be laid off proportional to the sqvare roots of the numbers
Fig.
-1
17.
01234567
-Method of constructing a double scale of squares or of
square roots.
0, 1, 2, 3, . . . that is, the square root of any length, when
laid off on OB, and marked with the symbol of the original length,
will he opposite the square root of that number on OA.
No difficulty need be experienced in carrying out the actual con-
struction of double scales representing algebraic relations, either by
use of a table of numerical values of the function or by means of
graphical construction. As a less laborious method of graphically
expressing functional relations will be explained in the next chapter,
the matter of double scales will not be discussed further at this place .
CHAPTER II
RECTANGULAR COORDINATES AND THE STRAIGHT
LINE
11. Statistical Graphs. Prom work in elementary algebra the
student is familiar with the construction of statistical graphs
simUar to Figs. 18 and 19. The student should carefully study
the construction of these two graphs. In Fig. 18, the point at
the center of any small circle represents the maximum temperature
(or the minimum temperature) on a particular day. This circle
is joined by a straight hne to the circle representing the maximum
(or minimum) temperature on the next day, and so on. The
lines joining the circles enable the eye to foUow at a glance the
changes of temperature for the entire month. However, a point
on a line between two circles has no meaning, because a point
on the horizontal scale between two consecutive points has no
meaning, for of course there is but one daily maximum for each
day. The student should especially note that the ratio of the
distance on the horizontal scale representing days to the distance
on the vertical scale representing a chajige of one degree in
temperature is so chosen as to make the fluctuations in the
temperatures stand out prominently. In constructing statistical
graphs, the student should always choose this ratio so that the
graph will clearly convey its intended meaning.
Smooth curves are drawn through the plotted points of Fig. 19
(not straight lines as in Fig. 18) because in this case intermediate
points have meaning; they represent temperatures at various
times of the day.
Fig. 20 is a barograph, or autographic record of the atmospheric
pressure recorded November 24, 1907, during a balloon journey
from Frankfort to Marienburg in West Prussia. The zero of the
scale of pressure does not appear in the diagram. Note also
- that the scale of pressure is an inverted scale, increasing downward.
23
24 ELEMENTARY MATHEMATICAL ANALYSIS t§ll
::: ^$; :: :^4::::::: :::":
-~-t,,_|V
=5^ -,- ^,
...bl,(^---
,.^:!j
,,-"-" ,.^:I
"x. / \
___3,_.^r- 1
3.^^^^ -t-
[ :v, \
J 1 .::l
1 ' ^r 1
Mi___g====:::.._.e^_
^ '^6,1 g
1 P P5 - -
--X T 1 .
^> - II
-- i -- J -- "is
..^Z____Z 1 i-""
_>. V^ SH^g
) S-^^M "'
..^- / ^ is
1 v___. _,__.,.§ %°
:^ l_ < ^
/l.f-'-'. 1— g
-- ---.^ U —-J »
::::::::^;:::::;:::=;;:::::t:::::i^:::":
v___.e4--ll--
1
asss8§ggs»
-§
7|2
Si*
si
-- B
--5J
3g
■d
^isqusiqcj saajSaQ — siTHEjaamaj;
§11]
RECTANGULAR COORDINATES
25
The scale of time is an algebraic scale,, the zero of which may be
arbitrarily selected at any convenient point. The scale of pres-
sure is an arithmetical scale. The zero of the barometric scale
'i70
—
—
—
—
—
— ~
—
—
—
—
~~
'
»^
■%eo
*-=k^
■g
'
•S^"
&
—
—
—
—
—
•^
b-
—
—
—
h^
n
$ 40
—
—
~ •
=u=
^-
Zt
^
—
—
—
—
_
G)
,
i>-j
1
-^
1
*? 30
~
"
-_
r:^
rt:
—
~~
~
—
~
~
—
~
§
Houriy_Air Temperatures
at
Madison Wisconsin
May,14.i910
and
Oct.. 10. 1910
0) .
|io
0
2 4
3
8 10 12_ 2 4 _
i
8 10 1?
Til
ne
oil
iaV
Hours
1/
Fig. 19. — Hourly air temperatures.
corresponds to a perfect vacuum — no less pressure and hence,
in this case, no negative value exists.
Fig. 21 is a graphical time-table of certain passenger trains be-
tween Chicago and Minneapolis. The curves are not continuous.
s
g
-3
^
s
=1
Fig. 20. — Barograph taken during a balloon journey. The vertical
scale is atmospheric pressure in miUimeters of mercury.
as in the case of the barograph, but contain certain sudden jumps.
What is the meaning of these? What indicates the speed of the
trains? Where is the fastest track on this railroad? What
shows the meeting jxrint of trains?
26
ELEMENTARY MATHEMATICAL ANALYSIS [§11
If the diagram, Fig. 21, he wrapped around a vertical cylinder of
such size that the two midnight lines coincide, then each train
line may be traced through continuously from terminus to terminus.
Functions having this remarkable property are said to be peri-
odic. In the present case the trains run at the same time every
day, that is, periodically. In mathematical language, the po-
sition of the trains is said to be a periodic function of the time.
Chicago
£au Olaire
Menomoaie
Hudson
St Paal
MinneapolieJf
10
12 2
A.M. Noon P.M.
Fig. 21. — Graphical time table of certain passenger, trains between
Chicago and Minneapolis.
Fig. 22 represents the fluctuation of the elevation of the ground-
water at a certain point near the sea coast on Long Island. The
fluctuations are primarily due to the tidal wave in the near-by
ocean. The curve is continuous. Is the curve periodic? What
indicates the rate of change in the elevation of the ground- water?
When is the elevation changing most rapidly? When is it
changing most slowly?
Fig. 23 represents the functional relation between the amount of
a domestic money order and the fee. This is an excellent illustra-
§12]
RECTANGULAR COORDINATES
27
tion of a discontinuous function. On account of the sudden
jumps in the values of the fee, the fee, as explained in the preceding
chapter, is said to be a discontinuous function of the amount of
the order.
Fig. 22. — Upper curve, elevation of water in a well on Long Island.
Lower curve, elevation of water in the nearby ocean.
12. Suggestions on the Construction of Graphs. Two kinds of
rectangular coordinate paper have been prepared for use with this
book. Form Ml is ruled in centimeters and fifths. Form M2
is ruled without major divisions in uniform 1/5-inch intervals.
It is a mistake to assume that more accurate work can be done on
finely ruled than on more coarsely ruled squared paper. Quite the
28
ELEMENTARY MATHEMATICAL ANALYSIS [§12
contrary is the case. Paper ruled to 1/20-inoh intervals does not per-
mit interpolation within the small intervals while paper ruled to 1/10
or 1/5-inch intervals permits accurate interpolation to one-tenth of the
smallest interval. Form Ml is ruled to 2-mm. intervals, and is fine
enough for any work. The centimeter unit has the very considerable
advantage of permitting twenty of the units within the width of an
ordinary; sheet of letter paper (SJ X 11 inches) while seven is the
largest number of inch units available on such paper.
In order to secure satisfactory results, the student must recog-
nize that there are several varieties of statistical graphs, and that
each sort requires appropriate treatment.
-50
MO 1
-30
r-20
HlO
1. 1. .Ill
j_
_L
J.
_!_
I
_L
J_
10
90
20 80 40 50 60 70 * 80
Amount of the Money Order in Dollars
Fig. 23. — The graph of a discontinuous function.
1. It is possible to make a useful graph when only one variable
is given. Thus Table I gives the ultimate tensile strength of
various materials.
A graph showing these results is given in Fig. 24. There are
two practical ways of showing the numerical values pertaining
to each material, both of which are indicated in the diagram;
either rectangles of appropriate height may be erected opposite the
name of each material, or points marked by circles, dots or crosses
may be located at the appropriate height. It is obvious in this
case that a smooth curve should not be drawn through these points
— such a curve would be quite meaningless. In this case there
§12] RECTANGULAR COORDINATES 29
Table I. — Ultimate Tensile Strength of Various Materials
Material
Tensile strength,
tons per square inch
Hard steel
50.0
30.0
25.0
21.5
16.0
12.0
11.0
10.0
5.0
Wrought iron
Drawn brass
Cast brass
Timber, with grain
are not two scales, but merely the single vertical scale. The hori-
zontal axis bears merely the names of the different materials
and has no numerical or quantitative signifioance. The result
is obviously not the graph of a function, for there are not two
variables, but only one. The graph is merely a convenient ex-
pression for certain discrete and independent results arranged
in order of descending magnitude.
2. It is possible to have a graph involving two variables in
which it is either impossible or undesirable to represent the graph
by a continuous curve or line. For example. Fig. 18 is a graph
representing the maximum temperature on each day of a certain
month. Because there is only one maximum temperature on
each day, the value corresponding to this should be shown by an
appropriate rectangle, or by marking a point by a circle, or by a
dot or cross, as in the preceding case. A continuous curve
through these points has no meaning. The horizontal scale may
be marked by the names of the days of the week or by numbers,
but in either case the horizontal line is a true scale, as it cor-
responds to the lapse of the variable time. Sometimes, as in Fig.
18, graphs of this kind are represented by marking the appropriate
points by dots or circles and then connecting the successive points
by straight lines. These lines have no special meaning in such
a case, but they aid the eye in following the succession of separate
points.
30
ELEMENTARY MATHEMATICAL ANALYSIS [§12
If a graph be made of the noonday temperatures of each day
of the same month referred to in Fig. 18, one of the same methods
indicated above would be used to represent the results; that is,
either rectangles, marked points, or marked points joined by lines.
Although a smooth curve drawn through the known points would
have a meaning (if correct), it is obvious that the noonday
temperatures alone are not sufficient for determining its form.
In all such cases a smooth curve should not be drawn.
3. If the data are reasonably sufficient, a smooth curve may,
and often should, be drawn through the known points. Thus if
the temperature be observed
every hour of the day and the re-
sults be plotted, a smooth curve
drawn carefully through the
plotted points will probably very
accurately represent the un-
known temperatures at interme-
diate times. The same may
safely be done in exercises (3)
and (4) below. In scientific
work it is desirable to mark by
circles or dots the values that
are actually given to distinguish
them from the intermediate
values "guessed" and repre-
sented by the smooth curve.
In addition to the above
suggestions, the student should
adhere to the following instruc-
tions :
4. Every graph should be
marked with suitable numerals
along both numerical scales.
5. Each scale of a statistical graph should bear in words a
description of the magnitude represented and the name of the
unit of measure used. These words should be printed in drafting
letters and not written in script.
•9
2
H
1
JS!a(\
Id-
1
',
\
i2<ift
\
H
-!
t-1
M
a
t-
OS
^
a in
■
"
1-
r"
-
~
n
.
_i
n D
Fig. 24. — Graph showing
tensile strength of certain struc-
tural materials.
§12]
RECTANGULAR COORDINATES
31
6. Each graph should bear a suitable title telling exactly what is
represented by the diagram.
7. The selection of the units for the horizontal and vertical
scales is an important practical matter in which common sense
must control. It is obvious that in the third exercise given
below 1 cm. =1 foot draft for the horizontal scale, and 1 cm.
= 100 tons for the vertical scale will be units suitable for use on
form Ml.
Further instruction in practical graphing is given in §33.
Exercises
1. Draw a statistical graph from the data given in the following
table. See Mg. 18. Represent the plotted points by small distinct
points, not by circles.
Maximum and Minimum Temperatures at Madison, Wisconsin,
FOR October, 1910
Date
Max. temp.,
Mm. temp.,
°F.
Date
Max. temp.,
"F.
Min. temp.,
"F.
1
68
55
17
81
53
2
71
^8
18
81
57
3
75
58
19
69
45
4
68
55
20
45
40
5
62
53
21
49
41
6
58
45
22
52
34
7
66
43
23
60
37
8
68
47
24
60
49
9
60
44
25
52
44
10
67
42
26
60
40
11
75
49
27
42
32
12
61
46
28
35
30
13
69
45
29
38
26
14
73
52
30
60
31
15
76
50
31
63
39
16
80
56
2. Draw a statistical graph for the data given in the following table.
See Fig. 19.
32
ELEMENTARY MATHEMATICAL ANALYSIS [§12
Hourly Air Temperatures at Madison, Wisconsin, Mat 14,
1910; October 10, 1910
Time
May 14,
1910,
temp., " F.
Oct. 10,
1910,
temp., " F.
Time
May 14,
1910,
temp., ° F.
Oct. 10,
1910,
temp., ° F.
1 a. m.
37
44
1 p. m.
58
65
2 a. m.
35
44
2 p. m.
61
66
3 a. m.
35
44
3 p. m.
63
67
4 a. m.
34
43
4 p. m.
62
67
5 a. m.
34
43
5 p. m.
62
65
6 a. m.
35
42
6 p. m.
61
62
7 a. m.
37
43
7 p. m.
69
57
8 a. m.
42
45
8 p. m.
66
65
9 a. m.
46
51
9 p. m.
53
55
10 a. m.
49
65
10 p. m.
50
66
11 a. m.
54
60
11 p. m.
47
62
12 a. m.
56
62
12 p. m.
45
51
3. At the following drafts a ship has the displacements stated :
Draft in feet, h ,
15
12
9
6.3
Displacement in tons, T
2096
1512
1018
586
Plot on squared paper. What are the displacements when the
drafts are 11 and 13 feet, respectively?
4. The following tests were made upon. a steam turbine generator:
Output in kilowatts, K.. .
1,190
995
745
498
247
Weight, pounds of steam
consumed per hour, W.
23,120
20,040
16,630
12,560
8,320
Plot on squared paper. What are the probable values of K when
W is 22,000 and also when W is 11,000?
6. The average temperature at Madison from records taken at 7
a. m. daily for 30 years is as. follows:
Jan. 1, 14.0.° F.
Feb. 1,15.1.
Mar. 1, 35.2.
Apr. 1, 40.0.
May 1, 53.9.
June 1, 63.^.
Make a suitable graph of these results on squared paper.
July
1, 67.5.
Aug.
1, 64.0.
Sept.
1, 55.4.
Oct.
1, 44.1.
Nov.
1, 30.0.
Dec.
1, 18.3.
§13]
RECTANGULAR COORDINATES
33
13. Rectangular Cobrdinates. Two intersecting algebraic
scales, with their zero points in common, may be used as a system
of latitude and longitude to locate any point in their plane. The
student should be familiar with the rudiments of this method
from the graphical work of elementary algebra. The scheme is
illustrated in its simplest form in Fig. 25, where one of the hori-
zontal lines of a sheet of squared paper has been selected as one
of the algebraic scales and one of the vertical lines of the squared
paper has been selected for the second algebraic scale. To locate
a given point in the plane it is merely necessary to give, in a
Y
4
'■7
1
Pr
(
iVzZH
)
Pi
(
-3,
2)
?,
II
I
X'
D
X
-:
-2
-
0
1
2
3
i
5
.^
P>
(-
2.-
1)
-?
I]
I
IV
-3
Y
Pi
(2,-S
)
^
Fig. 25. — Rectangular coordinates.
suitable unit of measiu'e (as centimeter, inch, etc.), the distance
of -the point to the right or left of the vertical scale and its distance
above or below the horizontal scale. Thus the point Pi, in Fig.
25, is 2j units to the right and 3j units above the standard
scales. P2 is 3 units to the left and 2 units above the standard
scales, etc. Of course these directions are to be given in mathe-
matics by the use of the signs "-)-" and " — " of the algebraic
scales, and not by the use of the words "right" or "left," "up"
or "down." The above scheme corresponds to the location of a
place on the earth's surface by giving its angular distance in
3
34 ELEMENTARY MATHEMATICAL ANALYSIS [§13
degrees of longitude east or west of the standard meridian, and
also by giving its angular distance in degrees of latitude north
or south of the equator.
The sort of latitude and longitude that is set up in the manner
described above is known in mathematics as a system of rectangu-
lar coordinates. It has become customary to letter one of the
scales XX', called the X-axis, and to letter the other YY', called
the Y-axis. In the standard case these are drawn to the right
and left, and up and down, respectively, as shown in Fig. 25.
The distance of any point from the F-axis, measured parallel to
the X-axis, is called the abscissa of the point. The distance of
any point from the X-axis, measured parallel to the F-axis, is
called the ordinate of the point. Collectively, the abscissa and
ordinate are spoken of as the coordinates of the point. Abscissa
corresponds to the longitude and ordinate corresponds to the
latitude of the point, referred to the X-axis as equator, and to
the F-axis as standard meridian. In the standard case, abscissas
measured to the right of YY' are reckoned positive, those to the
left, negative. Ordinates measured up are reckoned positive,
those measured down, negative.
Rectangular coordinates are frequently called Cartesian co-
ordinates, because they were first introduced into mathematics
by Ren6 Descartes (1596-1650).
The point of intersection of the axes is lettered 0 and is called
the origin. The four quadrants, XOY, YOX', X'OY', Y'OX,
are called the first, second, third, and fourth quadrants, respectively.
A point is designated by writing its abscissa and ordinate in a
parenthesis and in this order: Thus, (3, 4) means the point
whose abscissa is 3 and whose ordinate is 4. Likewise (—3, 4)
means the point whose abscissa is (—3) and whose ordinate is
(+4).
Abscissas are usually represented by the letter x and ordinates
are usually represented by the letter y. Thus the point whose
abscissa is 3 and whose ordinate is 4, may be described as the
point (3, 4), or equally well as the point x = 3, y — 4.
Unless the contrary is expUcitly stated, the scales of the eo-
■ ordinate axes are assumed to be straight and uniform and to inter-
sect at right angles. Exceptions to this are not uncommon.
§14] RECTANGULAR COORDINATES 35
Exercises
On suitable squared paper, select and mark a horizontal line as the
X-axis (or axis of abscissas) and select and mark a vertical line as the
F-axis (or axis of ordinates). Select and mark a suitable unit of
measure on each axis, for example 1 centimeter or 1/2 inch.
Then locate the points whose coordinates are given in the following
exercises.
1. Draw the coordinate axes on squared paper and locate the points
(3, 3), (2, 2), (1, 1), (0, 0), (-1, -1), (-2, -2), (-3, -3).
2. Draw the axes and locate the points (2, 3), (—2, 3), ( — 2, —3),
(2, -3).
3. Draw the coordinate axes and locate the points (5, 0), (4, 3),
(3, 4), (0, 5), (-3, 4), (-4, 3), (-5, 0), (-4, -3), (-3, -4), (0, -5),
(3, -4), (4, -3).
4. Draw suitable axes and locate the points ( — 3, —5), (—2, —3),
(-1, -1), (0, 1) (1, 3), (2, 5), (3, 7), (4, 9).
A brief way of describing a ^et of points is to place the abscissas and
ordinates in tabular form, indicating abscissas by the letter x and
indicating ordinates by the letter y, as follows :
a; I -3' -2 -10 1 2 3 4
y \ -5-3-113579
14. Mathematical, or Non-statistical Graphs. — Instead of the
expressions "abscissa of a ■point," or "ordinate of a point," it has be-
come usual to speak merely of the "x of a point," or of the "y of
a point," since these distances are conventionally represented by
the letters x and y, respectively. If we impose certain conditions
upon X and y, then it will be found that we have, by that very fact,
restricted the possible points of the plane located by them to a
certain array, or set of points, and that all other points of the
plane fail to satisfy the conditions or restrictions imposed.
It is obvious that the command, "Find the place whose latitude
equals its longitude," does not restrict or confine a person to a par-
ticular place or point. The places satifying this condition are
unlimited in number. We indicate all such points by drawing
a line bisecting the angles of the first and third quadrants; at all
points on this line latitude equals longitude. We speak of this
line as the locus of the point satisfying the conditions. We might
describe the same locus by saying "the y of each point of the
36
ELEMENTARY MATHEMATICAL ANALYSIS [§14
locus equals the x" or, with the maximum brevity, simply, write
the equation "y =; x." The equation "y = x" is called the
equation of the locus, and the line is called the locus of the
equation.
It is of the utmost importance to be able readily to interpret any
condition imposed upon, or, what is the same thing, any relation
between variables, when these are given in words. It will greatly
aid the beginner in mastering the concept of what is meant by the
term fimction if he will try to think of the meaning in words of the
relations commonly given by equations, and vice versa. The
very elegance and brevity of the mathematical expression of rela-
tions by means of equations, tends to make work with them formal
y
"h
i
2
^
/
/
1
/■
X'
0
/
X
1 D
/ ^'
Fig. 26. — The straight line y = x.
Fig. 27.— The straight line
y = 2x.
and mechanical unless care is taken by the beginner to express in
words the ideas and relations so briefly expressed by the equa-
tions. Unless expressed in words, the ideas are liable not to be
expressed at all.
The equation of a curve is an equation satisfied by the co-
ordinates of every point of the curve and by the coordinates of no
other point.
The graph of an equation is the locus of a point whose coordi-
nates satisfy the equation.
Illustration I. Find the equation of the Une of Fig. 26. The Une
§14]
RECTANGULAR COORDINATES
37
of this figure states that the y of any point of the line equals the x of
that point. Hence the equation of the line isy = x.
Illustration 2. The line of Fig. 27 is drawn through the origin and
the point (1, 2). Find the equation of the line. Let OB and DP be
the abscissa and ordinate, respectively, of any point on the line. Then
from similar triangles OPD and OPil, DP:OD = 2 : 1, or y : a; = 2: 1,
or y = 2x, which is the equation of the line.
Exercises
What is its
What is
What is
1. Draw a hne through the origin and the point (1, 3).
equation?
2. Draw a line through the origin and the point (1, -f)-
its equation?
3. Draw a line through the origin and the point (1, —1).
its equation? Draw a line through the
origin and the point (1, —2). What is its
equation?
4. Draw loci for the following and show
that each locus is a straight line passing
through the origin: (a) The ordinate of
any point of a certain locus is twice its
abscissa; (b) the x of every point of a cer-
tain locus is half its y; (c) the yoia. point is
1/3 of its x; (d) a point moves in such a
way that its latitude is always treble its
longitude; (e) the sum of the latitude and
longitude of a point is zero; (/) a point
moves so that the difference in its latitude and longitude is always
zero.
Hint: In part (a) let Pi (Fig. 28) be any point on the locus and
let Pi be any second point on the locus.
Draw OPi and OPi] draw PiDi and P2D2 perpendicular to OX. By
the conditions of the problem PiDi = 20Di and P2D2 = 20Di.
Hence
PiDi _P,D2
ODi OD2'
and the triangles OPiDi and OP2D2 are similar. Then the angles
PiODi and P2OD2 are equal. Hence OPi and OP2 coincide in direction
and 0, Pi, and P2, are upon a straight line.
5. Draw the locus: Beginning at the point (1, 2), a point moves so
Fig. 28. — Diagram
for exercise 4 (o) §14 and
, exercise 3 §15.
38
ELEMENTARY MATHEMATICAL ANALYSIS [§15
that its gain ia latitude is always twice as great as its gain in longitude.
Show that the locus is a straight line.
6. A point moves so that its latitude is always greater by 2 units
than three times its longitude. Write the equation of the locus and
construct- Show that the locus is a straight line.
15. Slope. The slope of a straight line is defined to be the
change in y for an increase in x equal to 1. It will be represented
in this book by the letter to. In Fig. 26 the line OP has slope 1 and
in Fig. 27 the line OP
has slope 2. Also in
Fig. 29 the line A has
the slope to = 1.5, for
it is seen that at any
point of the linb the
ordinate y gains 1.5
units for an increase
of 1 in X. The line B,
parallel to the line A,
is also seen to have the
slope equal to 1.5.
The equation of the
line A is obviously y
= 1.5a;. In the same
figure the slope of the
line C is ( — 2), for at
any point of this line
the ordinate 2/ decreases
2 units for an increase
in X equal to 1. The equation of the line C is obviously
y = —2x. Line D, parallel to line C, also has slope ( — 2).
If h be the change in y for an increase of x equal to k, then the
slope TO is the ratio h/k. Hence the practical method of determin-
ing the slope of a line drawn upon squared paper is: Select two
convenient points on the line rather far apart, and divide the change
in y by the increase in x.
The technical word slope differs from the word slope or slant in
common language only in the fact that slope, in its technical use,
is always expressed as the ratio of two algebraic numbers. In
Y
U
C
1
s
B
A
\
\
/
/
\
I
\
m
.-
2
4n
= 1
■y
\
\
/
/
t
m\
B-
2\
(^
/
1
/
\
\,
/
\
\
->
/
^m
= +
1..
\,
\
/
/
1
\
\
/
/
\,
/
\;
^m
-1
.5
X
/
^
/
X
-5
-4
-3
-2
/
i^-l
\
/
\
1
2
3
4
6
/
\
-1
\
m
--
2
/
/
V
\
'
/
\
\
/
t
\
s
/
/
-9
\
\
/
/
V
/
f
-\
\
/
/
\
\
/
/
-5
Y
s
..S
Fig. 29.-
-Lines of slope (1 . 5) and of slope
(-2).
§16] RECTANGULAR COORDINATES 39
common language we speak of a "slope of 1 in 10," or a "grade
of 50 feet per mile," etc. In mathematics the equivalents are
"slope = 1/10," "slope = 50/5280," etc.
As already indicated, the definition of slope requires us to speak
in mathematics of positive slope and negative slope. A line of posi-
tive slope extends upward with respect to the standard direction
OX and a line of negative slope extends downward with reference
toOZ.
In a similar way we may speak of the slope of any curve at a
given point on the curve, meaning thereby the slope of the tangent
line drawn to the curve at that point.
Exercises
1. Give the slopes of the lines in exercises 1 to 6 of the preceding set
of exercises.
2. Draw y = x; y = 2x; y = Zx; y = -^-t y = ^, y = ^, y = — 2x;
y = — 3z; y = Ox.
3. Prove that y = mx always represents a straight line, no matter
what value m may have. Hint: Make use of Fig. 28.
16. Equation of a Straight Line. Intercepts. — In Fig. 30, the
line MN expresses that the ordinate y is, for all points on the line,
always 3 times the abscissa x, or it says that y = 3a;. The line
HK is 2 units higher than MN, so that it states that "2/ is 2 more
than 3a;." Thus the line HK has the equation y = Zx + 2. In
Fig. 29 the line 5 is 2 units higher than the line y = 1.5a;, hence
its equation is y = 1.5x -{- 2. The line D is 2 units lower than
the line C, whose equation is y = —2x, hence the equation of
Disy = -2x - 2.
In general, since y = mx is always a straight line,^ then y =
ma; + 6 is a straight line, for the y of this locus is merely, in each
case, the y of the former increased by the constant amount 6 (which
may, of course, be positive or negative). Therefore, y — mx + 6
is a line parallel to y = mx. The line y = mx + 6 is 6 units
higher than, or above, the line y = mx ii b stands for a positive
number and the line y = mx + b is b units lower than, or below
the line y = mx if 6 stands for a negative number. The distance
1 See exercise 3, §15, above.
40
ELEMENTARY MATHEMATICAL ANALYSIS l§16
OB (Fig. 30) is equal to b, and is called the Y-intercept of the
graph. The distance OA is equal to — b/m, for it is the value
of X obtained from the equation when y is given the value zero.
It is called the X-intercept of the locus. The equation
7 = mz -|- b is called the common equation of the straight line.
X'
1
K
a 1
2
r
s /
1
B
1 /
T—r
A /
0
-1
: 3
1
1
'
/
2 /
-9.
L
/
-S
1
J
y'
-4
X
Fig. 30. — Intercepts. MN is the line j/ = 3x; UK is the line y =
3s + 2 ; OB, the F-intercept of HK, is equal to 2 ; OA, the Z-intercept,
is equal to —2/3.
IllustraMon 1. Sketch the line y = 2s + 3.
This line is 3 units higher than the line y = 2x. Hence through the
point (0, 3) on the 7-axis, draw a line of slope 2, which is the required
line. I
Illustration 2. Sketch the line ?/ = — 2.r — 1.
§16] RECTANGULAR COORDINATES 41
The line is 1 unit lower than the line y = —2x. Hence through
the point (0, —1) on the F-axis, draw a line of slope (—2). The line
lies halfway between lines C and D (Fig. 29).
Illustration 3. Draw the line whose equation is 4a! — 2?/ — 3 = 0.
Solve the equation for y by transposing the terms 4a; and (—3) to
the right member and dividing both members by ( — 2), then
y = 2x-i.
Hence through the point (0, — f ) on the y-axis draw a line of slope 2.
Exercises
1. Sketch, from inspection of the equations, the lines given by:
(a) y ^ X. (d) y = X + 3.
(b) y = X + 1. (e) y = X - 1.
(c) y =x + 2. if) y =x -2.
2. Sketch, from inspection of the equations, the lines given by:
(a) y = ix. (/) y = -\x.
(6) y = ix. (g) y = -x.
(c) y = X. (h) 2/ = -2a;.
{d) y = 2x. (i) y = -3x.
- (e) 2/ = 3a;. (j) y = y/2 x.
3. Sketch the lines given by the equations:
(a) a; = 3. {d) y = 1. (s) 2/ = 0.
(6) a; = 6. (e) y = 5. (A) x = 0.
(c) X 2. t/) y = -3. (i)a;2 = 4.
4. Sketch and name the slope and F-intercept in each of the
following:
(o) 2/ = a; + 1. (/) 2/ = 3x + 4.
(6) 2/ = ia; + 1. (?) 2/ = a; - 6.
(c) 2/ = -2a; + 4. Qi) 2/ = fx + 8.
{d) 2/ = 6x + 3. (i) 2/ = -3x + 4.
(e) y = —Sx — 2. 0') 2/ = — ia; — 3.
5. Give the slope and F-intercept for each of the following :
(a) y =2x + Z. (/) 3y - 6x = 12.
(6) y = 3x -2. (9) y +x = 1.
(c) 2/ = -3x - 1. I^h) 3y'+ 2x = 7.
(d) 2/ = 5x— 6. (i) X — ^ = 6.
(e) 22/ = X + 4. (j) X - 22/ = I.
42 ELEMENTARY MATHEMATICAL ANALYSIS [§16
6. Find the X-intercept and the }'-intercept for each of the
following :
(.a) Sx - 2y = 5. (e) y - 2x ~ 6 ,= 0.
(6) 2x + y =Q. if) 2y + 3x + 5 = 0.
(c) X -y = 7. (g) X + y + I = 0.
(d) y -3x = 5. (h) 5y - 3x + 10 = 0.
7. Name the slope and the F-intercept in each of the following;
(a) 2y =x + 4:. {f) ix = 3y - 6.
(6) y -2x-3 =0. (ff) Wx-y = 7.
(c) y + fa; + J = 0. (h) ax -\=y.
(d) 2y -\-3x = 4. (i) ax + by = c.
(e) 2x -3y = 6. 0') x/a + y/b = 1.
8. What is the equation of the X-axis? Of the K-axis?
9. What ia the equation of a line parallel to the X-axis 4 units
above? 3 units above? 10 units below? 60 units below?
10. What is the equation of a line parallel to the F-axis 3 units
to the right? 20 units to the right? 7 units to the left? 100 units
to the left?
11. Plot the following pairs of points on squared paper, and draw
the line determined by each pair:
(a) (-1, 3) and (5, -6)
(6) (-2, -5) and (3, 4)
(c) (1, 1) and (7, -8).
Find the slope and, by means of similar triangles, find the F-intercept
of each line. Write the equation of each line by replacing m and 6
in y = mx -|- 6 by the values found for slope and intercept. Test
the correctness of the equations by substituting for x and y the co-
ordinates of the given points.
12. A head of 100 feet of water causes a pressure at the bottom of
43.4 pounds per square inch. Draw a graph showing the relation
between head and pressure, for all heads of water from 0 to 200 feet.
StTGGESTiON: There are several ways of proceeding. Let pounds
per square inch be represented by abscissas or x, and feet of water be
represented by ordinates or y. Since negative numbers are not in-
volved ia this exercise, the origin may be taken at or near the lower
left corner of the squared paper. Draw a line through the points
(0, 0) and (86.8, 200). This will be the required graph. Otherwise
100
produce the equation y = 73-72; from the proportion x:y= 43.4 : 100
§16] RECTANGULAR COORDINATES 43
and then draw the graph from the fact that the latitude is always
TK~r of the longitude. In drawing this graph let 2 centimeters on the
X-axis represent 10 units, and 1 centimeter on the F-axis represent
10 units. Be sure that the scales are numbered and labelled in
accordance with suggestions (4), (5), and (6) of §12. On the X-axis
mafk only the points corresponding to hundreds of pounds, and on'
the y-axis mark only the points corresponding to tens of feet.
13. From the straight line drawn in exercise 12, find pressure meas-
ured in pounds corresponding to 13.1, 112.6, 93.7, and 187.5 feet of
water.
14. From the straight line drawn in exercise 12, find the head in
feet of water corresponding to 1123, 178, and 89 pounds per square
inch.
16. A pressure of 1 pound per square inch is equivalent to a column
of 2:04 inches of mercury, or to one of 2.30 feet of water. Draw a
graph showing the relation between pressure expressed in feet of water
and pressure expressed in inches of mercury.
StrGGBSTiON: Let x = inches of mercury and y = feet of water.
First properly number and label the X-axis to express inches of mer-
cury and number and label the K-axis to express feet of water. Since
negative numbers are not involved in this exercise, the origin may be
taken at the lower left-hand corner of the squared paper. First locate
the point x = 2.04; y = 2.30 (which are the corresponding values
given by the problem) and draw a line through it and the origin. This
is the required locus since at all points we must have the proportion
x:y:: 2.04 : 2.30, which says that the ordinate of every point of
the locus is 2.30/2.04 times the abscissa of that point.
16. A certain mixture of concrete (in fact, the mixture 1:2:5) con-
tains 1.4 barrels of cement in a cubic yard of concrete. Draw a
graph showing the cost of cement per cubic yard of concrete for a
range of prices of cement from $0.80 to $2.00 per barrel.
Suggestion: Let x be the price per barrel of cement and y be the
cost of the cement in 1 cubic yard of concrete. Let 2 centimeters on
both vertical and horizontal scales represent 10 cents. Number only
the points representing multiples of 10 cents. Since that portion of
the graph near the origin, namely to the left of 0,.80 and below 1.12
will not be used, place the scales on the horizontal and vertical lines
passing through the point (0.80, 1.00) and place this point at or near
the lower left corner of the paper. The X- and F-axes will not
appear on the drawing.
17. Draw a graph showing the cost per cubic yard of concrete for
44 ELEMENTARY MATHEMATICAL ANALYSIS [§17
various prices of cement, provided $2.10 per yard must be added to
the results of example 16 to cover cost of sand and crushed stone.
18. Cast iron pipe, class A (manufactured for heads under 100 feet),
weighs, per foot of length: 4-inch, 20.0 pounds; 6-inch, 30.8 pounds;
8-inch, 42.9 pounds. Upon a single sheet of squared paper, construct
a graph for each size of pipe, showing the cost per foot for all variations
in market price between $20 and $40 per ton.
Suggestion: If the horizontal scale be selected to represent
•price per ton, the scale may begin at 20 and end at 40, as this covers
the range required by the problem. Therefore let 1 centimeter
represent $1.00. Since the range of prices is from 1 cent to 2 cents
per pound, the cost per foot will range from 20 cents to 40 cents for
4-inch pipe and from 42.9 cents to 85.8 cents for 8-inch pipe.
Hence for the vertical scale 10 cents may be represented by 2
centimeters. In this case the vertical scale may quite as well begin
at 0 cents instead of at 20 cents, as there is plenty of room on' the
paper.
19. Show that the shortest distance between y — mx and y = mx + 6
is not 6, but — ,
20. Pick out two pairs of parallel Unes in exercise 5, above . Pick
out a pair of parallel lines in exercise 4, above.
17. Line with Slope and One Point, or with Two Points Given. —
The equation of any line parallel to the F-axis is of the form x = a,
which is an equation in which the variable y does not appear. The
equation of all other lines may be written in the form
y = mx + h, j (1)
in which m is the slope of the line and h is the F-intercept. Two
important special cases are explained below.
Illustration 1. Find the equation of the line of slope 4 which passes
through the point (2, 3).
Since to = 4, equation (1) becomes
y =4x + b. (2)
Replacing a; by 2 and y by 3, we get
3 = 8 -F 6, or 6 = -6.
Hence the equation of the line of slope 4 passing through (2, 3) is
y = 4x - 5, (3)
: §17] RECTANGULAR COORDINATES 45
Illustration 2. Find the equation of the line passing through the
pomts (2, 3) and (4, 1).
Substituting the given values of x and y in equation (1) we have
3 = 2m + 6
1 = 4w + 6.
Solving these equations for the two unknown numbers, m and 6, we
find
6 = 5
m = —1,
so that the equation of the line passing through the given points is
2/ = -X + 5.
In like manner the equation of a line passing through any two given
points may be found. In geometry we learned that two points de-
termine a straight line, and in the present problem the coordinates
of two given points are necessary and sufficient for the determination
of the equation of the line.
Exercises
1. Find the equation of the line determined by each of the following
conditions:
(a) Passes through (2, 5) and has slope 3.
(6) Passes through ( — 2, 6) and has slope — 2^.
(c) Passes through (4, —1) and has slope 7.41.
2. Find the equation of the line determined by each of the following
pairs of points:
(o) (3, 2) and (1, 5) (c) (4, 6) and (3, -2)
(6) (1, 2) and (- 2, 6) (d) (0, 0) and (-2, -3)
3. Show that the equation of the line passing through (a, 0) and
(0, 6) niay be written in the form
-+! = !■
a 0
4. Find the equations of the three sides of the triangle whose ver-
tices are the points (0, 3), (2, 4), and (5, 9).
In each of the following exercises certain observed data are tabu-
lated which will be found in each case to give points lying on a straight
line. The law connecting y and x must then be of the form
y = mx + b.
46 ELEMENTARY MATHEMATICAL ANALYSIS [§17
6. Find the law connecting y and x when the following correspond-
ing values are given :
X I 10 25 54 72
17 47 105 141
Hint: In plotting, take the origin at the lower left corner of the
squared paper. Let 2 centimeters represent 10 units on the X-axis
and 1 centimeter represent 10 units on the K-axis. To find the slope
divide the change in ?/ or (141 — 17) by the increase in a; or (72 — 10)
which gives 2. Find F-intercept by method of Illustration 1.
6. Find the law connecting x and y from the following data :
X I 12.0 15.3 17.8 19.0
y \ 24.2 29.0 32.6 34.2
Hint: Take origin near left lower corner of the squared paper and
let 4 centimeters equal 10 units on each axis.
7. L is the latent heat of steam at a temperature i° C Find a
simple formula giving L in terms of t.
« I 75 90 100 115 125
L I 554 544 536 526 519
Hint: Call the lower left corner of the squared paper the point
t = 75, L = 500. Let 1 centimeter = 5 units on each axis.
8. V is the volume in cubic centimeters of a certain weight of gas
at temperature t° C, the pressure being constant. Find the law
connecting V and t.
t I 27.0 33.0 40.0 55.0 68.0
V I 109.9 112.0 114.7 120.1 125
Hint: Call the lower left corner of the squared paper the point
t = 25, V = 100. Let 2 centimeters equal 5 units on the t scale and
1 centimeter equal 1 unit on the V scale.
9. I feet is the length of an iron bar under a pulling stress of W tons.
Find the law connecting I and W.
W I 0 1 1^8 3^2 4^2 6.0
I I 10 10.005 10.010 10.0175 10.0225 10.0325
Hint: Call the lower left comer of the squared paper the point
TF = 0, Z'= 10. Let 2 centimeters = 1 unit on the W scale and 1
qentimpter = 0.005 unit? on the I scale.
§17] RECTANGULAR COORDINATES 47
10. The following table gives the draw-bar pull in pounds (P) of
an electric locomotive in terms of the current consumed (A). Find
an approximately correct algebraic formula giving A for any value
of P. Find the current required for a pull of 2000 pounds.
P I 400 800 1370 1600 2080 2400 ,
A I 65 86 106 116 137 150
Hint: Call the lower left corner of the squared paper A = 50, P =
400. Let 2 centimeters equal 10 units on the A scale and 1 centimeter
equal 100 units on the P scale. /
Exercises 5-10 above are taken from Saxelby's "Practical Mathe-
matics," Longmans, Green and Co., New York, 1905.
CHAPTER III
THE POWER FUNCTION
18. Definition of the Power Function. The algebraic function
consisting of a single power of the variable, such for example as
the functions x'^, x', 1/x, 1/x'', a;^''', etc., stand next to the linear
function of a single variable, mx + b, in fundamental impor-
tance. The function a;" is known as the power function of x.
19. The Graph of x^. The variable part of many functions of
practical importance is the square of a given variable. Thus the
area of a circle depends upon the square of the radius; the distance
traversed by a falling body depends upon the square of the
elapsed time; the pressure upon a flat surface exposed directly
to the wind depends upon the square of the velocity of the
wind; the heat generated in an electric current in a given time
depends upon the square of the number of amperes of current,
etc. Each of these relations is expressed by an equation of the
form y = ax^, in which x stands for the number of units in one of
the variable quantities (radius of the circle, time of fall, velocity of
the wind, amperes of current, respectively, in the above named
cases) and in which y stands for the other variable dependent
upon these. The number a is a constant which has a value
suitable to each particular problem, but in general is not the same
constant in different problems. Thus, if y be taken as the area of
a circle, y = irx^, in which x is the radius measured in feet or
inches, etc., and y is measured in square feet or square inches,
etc. ; or if s is the distance in feet traversed by a falling body,
thens = 16. li'', where i stands for the elapsed time in seconds.
In one case the value of the constant a is 3.1416 and in the other
its value is 16.1.
Let us first graph the abstract law or equation y = x^, in which
a concrete meaning is not assumed for the variables x and y but
48
§20]
THE POWER FUNCTION
49
in which both are thought of as abstract variables,
suitable table of values for x and x^ as follows:
First form a
X
a;2 or y
X
x^ or y
- 3.0
9.00
0.2
0.04
- 2.5
6.25
0.4
0.16
- 2.0
4.00
0.6
0.36
- 1.8
3.24
0.8
0.64
- 1.6
2.56
1.0
1.00
- 1.4
1.96
1.2
1.44
- 1.2
1.44
1.4
1.96
- 1.0
1.00
1.6
2.56
- 0.8
0.64
1.8
3.24
- 0.6
0.36
2.0
4.00
- 0.4
0.16
2.5
6.26
- 0.2
0.04
3.0
9.00
0.0
0.00
Here we have a series of pairs of values of x and y which are asso-
ciated by the relation y — x'^. Using the x of each pair of values
as abscissa with its corresponding y there can be located as many-
points as there are pairs of values in the table, and the array of
points thus marked may be connected by a freely drawn curve.
To ,draw the curve upon coordinate paper, form Ml, the origin
may be. taken near the lower mid-point of the sheet, and 2
centimeters used as the unit of measure for x and y. If the points
given by the pairs of values are not located fairly close together,
it is obvious that a smooth curve cannot be satisfactorily sketched
between the points until intermediate points are located by using
intermediate values of x in forming the table of values. The
student should think of the curve as extending indefinitely
beyond the limits of the sheet of paper used; the entire locus
consists of the part actually drawn and of the endless portions
that must be followed in imagination beyond the range of the
paper. If the graph of i/ = a;^ be folded about the F-axis, OY,
it will be noted at once that the left and right portions of the
curve will exactly coincide. The student will explain the reason
for this fact.
20. Parabolic Curves. The equations y = x, y — x', y = xi,
y = x^ should be graphed by the student on a single sheet'of coor-
60 ELEMENTARY MATHEMATICAL ANALYSIS [§20
dinate paper, using 2 centimeters as the unit of measure in each
case.^ Table II may be used to save numerical computation in
the construction of the graphs of these power functions. As in
the case oi y = x^, a smooth curve should be sketched free-hand
t
L
1
1
1
1
1
I 5
Y I
1
1
1
1
1
1/'
-\
|---
1
1
1
1
4
^
/..
Vf
1 ^
--f--
--\-'f
4--
1 2
--f-
-A---
I\
1 1
--J--
/ j
X'
— -|--
\L..
---]/-
— I
X
2 1
1 1 0
1
Y' j
1 ! 2
Fig. 31. — The parabola y = x'.
through the points located by means of the table of values, and
intermediate values of x and y should be computed when doubt
exists in the mind of the student concerning the course of the
curve between any two points.
1 Wlieii.aquBred paper ruled in inches is used instead of form M 1, one inch should
b« taken as th« unit of measure.
§20]
THE POWER FUNCTION
Table II
51
X
x^'
X'
Vi
</x
^%
1/x
1/x^
0.2
0.04
0.008
0.447
0.585
0.089
5.000
25.000
0.4
0.16
0.064
0.632
0.737
0.252
2.500
6.250
0.6
0.36
0.216
0.775
0.843
0.46S
1.667
2.778
0.8
0.64
0.512
0.894
0.928.
0.715
1.250
1.563
1.0
1. 00
1.000
1.000
1.000
1.000
1.000
1.000
1.2
1.44
1.72S
1.095
1.063
1.312
0.8333
0.6944
1.4
1.96
2.744
1.183
1.119
1.657
0.7143
0.5102
1.6
2.56
4.096
1.265
1.170
2.034
0.6250
0.3906
1.8
3.24
5.832
1.342
1.216
2.415
0.6556
0.3086
2.0
4.00
8.000
1.414
1.260
2.828
0.5000
0.2600
2.^
4.84
10.65
1.483
1.301
3.263
0.4646
0.2066
2.4
5.76
13.82
1.549
1.339
3.717
0.4167
0.1736
2.6
6.76
17.58
1.612
1.375
4.193
0.3846
0.1479
2.8
7.84
21.95
1.673
1.409
4.685
0.3571
0.1276
3.0
9.00
27.00
1.732
1.442
5.196
0.3333
0.1111
3.2
10.24
32.77
1-.789
1.474
5.724
0.3125
0.0977
3.4
11.56
39.30
1.844
1.504
6.269
0.2941
0.0865
3.6
12.96
46.66
1.897
1.533
6.831
0.2778
0.0772
3.8
14.44
54.87
1.949
1.560
7.407
0.2632
0.0693
4.0
16.00
64.00
2.000
1.587
8.000
0.2500
0.0625
4.2
17.64
74.09"
2.049
1.613
8.608
0.2381
0.0567
4.4
19.36
85.18
2.098
1.639
9.229
0.2273
0.0517
4.6
21.16
97.34
2.145
1.663
9.866
0.2174
0.0473
4.8
23.04
110.6
2.191
1.687
10.42
0.2083
0.0434
5.0
25.00
125.0
2.236
1.710
11.18
0.2000
0.0400
5.2
27.04
20.16
140.6
2.280
1.732
11.85
0. 1923
0.0370
5.4
157.5
2.324
1.754
12.66
0.1862
0.0343
5.6
31.36
175.6
2.366
1.776
13.25
0.1786
0.0319
5.8
33.64
195.1
2.408
1.797
13.97
0.1724
0.0297
6.0
36.00
216.0
2.449
1.817
14.70
0.1667
0.0278
6.2
38.44
238.3
2.490
1.837 :
15.44
0.1613
0.0260
6.4
40.96
262.1
2.530
1.867
16.19
0.1563
0.0244
6.6
43.56
287.5
2.569
1.876
16.96
0.1515.
0.0230
6.8
46.24
314.4
2.608
1.895
17.33
0.1471
0.0216
7.0
49.00
343.0
2.646
1.913
18.52
0.1429
0.0204
7.2
51.84
373.2
2.683
1.931
19.32
0.1389
0.0193
7.4
54.76
405.2
2.720
1.949
20.13
0.1351
0.0183
7.6
57.76
439.0
2.757
1.966
20.95
0.1316
0.0173
7.8
60.84
474.6
2.793
1.983
21.79
0.1282
0.0164
8.0
64.00
512.0
2.828
2.000
22.63
0.1260
0.0156
8.2
67.24
551.4
2.864
2.017
23.48
0.1220
0.0149
8.4
70.56
592.7
2.898
2.033
24.35
0.1190
0.0142
8.6
73.96
636.1
2.933
2.049
25.22
0.1163
0.0135
8.8
77.44
681.5
2.966
2.065
26.11
0.1136
0.0129
9.0
81.00\
729.0
3.000
2.080
27.00
0.1111
0.0123
9.2
84.64
778.7
3.033
2.095
27.91
0.1087
0.0118
9.4
88.36
830.6
3.066
2.110
28.82
0.1064
0.0113
9.6
92.16
884.7
3.098
2.125
29.74
0.1042
0.0109
9.8
96.04
941.2
3.130
2.140
30.68
0.1020
0.0104
10.0
100.00
1000.0
3.162
2.154
31.62
0.1000
0.0100
All of the graphs here considered have one important property
in common, .namely, they all pass through the points (0, 0) and
(1, 1). It is obvious that this property may be affirmed of any
52 ELEMENTARY MATHEMATICAL ANALYSIS [§21
curve of the class y = a;", if n is a positive number. These curves
are known collectively as curves of the parabolic family, or simply
parabolic curves. The curve y = x^ is called the parabola.
2/ = a;' is called the cubical parabola, y = xi is called the semi-
cubical parabola, etc. Curves for negative values of n do not pass
through the point (0, 0) and are otherwise quite distinct. They
are known as curves of the hyperbolic type, and will be discussed
later.
The student should cut patterns df the parabola, the cubical
parabola and the semi-cubical parabola out of heavy paper for
use in drawing these curves when required. Each pattern should
have drawn upon it either the X- or F-axis and one of the unit
lines to assist in properly adjusting . the pattern upon squared
paper.
21. Symmetry. — In geometry a distinction is made between two
kinds of symmetry of plane figures — symmetry with respect to a
line and symmetry with respect to a point. A plane figure is
symmetrical with respect to a given line if the two parts of the
figure exactly coincide when folded about that line. Thus the let-
ters M and W are each symmetrical with respect to a vertical line
drawn through the vertex of the middle angles. We have already
noted that y = x^ is symmetrical with respect to OY.
A plane figure is symmetrical with respect to a given point when
the figure remains unchanged if rotated 180° in its own plane about
an axis perpendicular to the plane at the given point. Thus the
letters N and Z are each symmetrical with respect to the mid-point
of their central line. The letters H and O are symmetrical both
with respect to lines and with respect to a point. Which sort of
symmetry is possessed by the curve y = a;'? Why?
Another definition of symmetry with respect to a point is per-
haps clearer than the one given by the above statement: A curve
is said to be symmetrical with respect to a given point 0 when all
lines drawn through the given point and terminated by the curve
are bisected at the point 0.
What kind of symmetry with respect to one of the coordinate
axes or to the origin (as the case may be) does the point (2, 3) bear
to the point (-2, 3)? To the point (-2,-3)? To the point
(2, -3)?
§22]
THE POWER FUNCTION
53
Note that symmetry of the first kind means that a plane figure is
unchanged when turned 180° about a certain line in its plane, and
l-r-Y
Fig. 32. — The parabolas «/ = a;" for n = 1, 2, 3, and 4.
that symmetry of the second kind means that a figure is unchanged
when turned 180° about a certain line perpendicular to its plane.
22. The curves in Figs. 31 to 34 are sketched from a limited
54
ELEMENTARY MATHEMATICAL ANALYSIS [§22
number of points only, but any number of additional values of x
and y may be tabulated and the accuracy, as well as the extent,
of the graph be made as great as desired. A number of graphs of
power functions are shown as they appear in the first quadrant
in Figs. 34 and 38. The student should explain how to draw the
portions of the curves lying in the other quadrants from the part
appearing in the first quadrant.
In the exercises in this book to "draw a curve" means to con-
THE POWER FUNCTION
55
struct the curve as accurately as possible from numerical or other
data. To "sketch a curve" means to produce an approximate or
less accurate representation of the curve, including therein its
characteristic properties, but without the use of extended numer-
ical data. Whenever possible, make use of the paper patterns
mentioned in §20.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 IJl 1.6 1.7 1.8 1.9 2.0
!Fig. 34. — Graph of the power function for n >0 (parabolic curves) in
the first quadrant.
Exercises
1. On coordinate paper draw the curves y — x', y = x^, y = x^,
y = x^, using 2 centimeters as the unit of measure. On the same
sheet draw the Unes x = +1, y = ±1, y — +x.
2. On coordinate paper sketch the curves x = y^, x = y^, x = yi,
X = y^. Compare with the curves of Exercise 1.
56
ELEMENTARY MATHEMATICAL ANALYSIS [§23
3. Sketch and discuss the curves y = xi, y = xi, y = xi. Can
any of these curves be drawn from patterns made from the
curves of Exercise 1? Why?
4. Draw the curve y^ = x*. Compare with the curve y = x^.
6. Name in each case the quadrants in which the curves of Exer-
cises 1-4 lie, and state the reasons why each curve exists in certain
quadrants and not in the other quadrants.
23. Discussion of the Parabolic Curves. — Draw the straight
ines X = 1, X = —l,y=l,y= —1 upon the same sheet upon
which a number of para-
■,/J;^. :^ I B bolic curves have been
drawn. These lines to-
gether with the coordi-
nate axes divide the plane
into a number of rect-
angular spaces. In Fig.
35 these spaces are shown
divided into two sets,
those represented by the
cross-hatching, and those
shown plain. The cross-
hatched rectangular spaces
'iontain the lines y = x
and y = —x and also all
curves of the parabolic
type. No parabolic curve
enters the rectangular strips
shown plain in Fig. 35.
The line y = x divides the spaces occupied by the parabolic
curves into equal portions. Why does the curve y = x' (in the
first quadrant) lie below this line in the interval a; = 0 to a; = 1,
but above it in the interval to the right of x = 1 ? On the other
hand, why does the curve y = xi, or y^ = x (in the first quad-
rant), lie above the line 2/ = a; in the interval a; = 0 to a; = 1 and
below y = xin the interval to the right of x = 1?
One part of the parabolic curve y = x" always lies in the first
quadrant. If n be an even integer, another part of the curve lies
in which quadrant? If n be an odd integer, the curve lies in which
quadrants?
Fig. 35. — The regions of the parabohc
and the hyperboUc curves. All parabolic
curves he within the cross-hatched
region. All hj^jerboUc curves he within
the region shown plain.
§23] THE POWER FUNCTION 57
If the exponent n of any power function be a positive fraction,
say m/r, the equation of the curve may be written y = x". If
in this case both m and r be odd, the curve lies in which quadrants?
If m be even and r be odd, the curve lies in which quadrants?
If m be odd and r be even, the curve lies in which quadrants?
If both m and r be even the curve lies in which quadrants?
A curve which is symmetrical to another curve with respect to a
line may figuratively be spoken of as the reflection or image of the
second curve in a mirror represented by the given line.
Exercises
Exercises 1-5 refer to curves in the first quadrant only.
3
1. The expressions x', x^, x', x* are numerically less than x for
values of x between 0 and 1. How is this fact shown in Fig. 34?
2. The expressions x^, x^, a;', s* are numerically greater than x for
all values of x numerically greater than unity. How is this fact
pictured in Fig. 34?
3. For values of x between 0 and 1, x* <x^ <x'' <x^ < x. For
values X > 1, X* > x^ > x^ > ^ > X. Explain how each of these
facts is expressed by the curves of Fig. 34.
2 i i 1
4. Show that the graphs y = x^, y = a;', y = x'l y = x^ are
the reflections of i/ = x^, y = x', y = x\ y = x\ in the line y = x.
6. Sketch on a single sheet of squared paper without tabulating
the numerical values, the following loci: y = x^", y = a;"', y = a;'"",
y = a;0.01
The following are to be discussed for all quadrants.
6. Sketch, without tabulating numerical values, the following loci
y' = x\ y* = x', y^ = x^, y' = x^, y^ = x^.
7. Show that y = —x' is the reflection oi y = x' in the X-axis.
8. Sketch the following; y = —x, y = —x^, y = —x', y' = — x^,
y^ — — x^.
9. A ball roUs down a smooth inclined plane making an angle of
45° with the horizontal. The distance s measured in feet along the
incline is given by the formula
s = 11.4*2
where t is time in seconds. Draw a, graph for this equation. Let
time t be represented by distances along the axis of abscissas and dis-
58
ELEMENTARY MATHEMATICAL ANALYSIS [§24
tance s along the axis of ordinates. Let 2 centimeters represent one
second on one axis and 10 feet on the other.
24. Hyperbolic Curves. Loci of equations of the form yx" = 1,
OT y = l/x", where n is positive, are called hyperbolic curves.
The fundamental curve xy = 1, or y = l/x is called the rec-
tangular hyperbola. Its graph is given in Fig. 36, but the curve
Fig. 36. — The hyperbolas y = x" ioi n = — 1, —2, and —3.
should be drawn independently by the student, using 2 centi-
meters as the unit of measure. Its relation to the X- and
y-axes is most characteristic. For a very small positive value of
X, the value of y is very large, and as x approaches 0, y increases
indefinitely. But the function is not defined for x = 0, for the
product xy cannot equal 1 if x be zero. For numerically small but
THE POWER FUNCTION
59
negative values of x, y is negative and numerically very large, and
becomes numerically larger as x approaches 0.
Instead of saying that "y increases in value without limit," it
is just as common to say "y becomes infinite;" in fact, "infinite"
is merely the Latin equivalent of "no limit." It is often written
2/ = 00 . This is a mere abbreviation for the longer expressions,
"y becomes infinite" or "y increases in value without limit."
The student must be cautioned that the symbol <» does not stand
Fig. 37. — The hyperbolas y = x" for n
-i, —i, —21 and —J
for a number, and that "y = oo" must not be interpreted in the
same way that "y = 5" is interpreted.
As X increases from numerically large negative values to 0,
y continually decreases and becomes negatively infinite (abbre-
viated 2/ = — 00 ). As a; decreases from numerically large positive
values to 0, y continually increases and becomes infinite. Thus,
in the neighborhood oi x = 0, y is discontinuous, and, in this case,
the discontinuity is called an infinite discontinuity.
60
ELEMENTARY MATHEMATICAL ANALYSIS [§24
0 0.1 0.2 0.3 0.4 O.S 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.1 1.6 1.6 1.7 1.8 1.9 2.0
Fig. 38. — Hsrperbolic curves in the first quadrant. y = l/s^-^o'is
the adiabatic curve for air.
On account of the symmetry in xy = l,]i we look upon x as a
function o^ y, all of the above statements may be repeated, merely
interchanging x and y wherever
they occur. Thus, there is an in-
finite discontinuity in x, as y
passes through the value 0.
The lines XX' and YY' which
these curves approach as near as'
we please, but never meet, are
called the asymptotes of the
hyperbola.
All other curves of the hyper-
bolic family, such as yx"^ = 1,
^2/" = 1, 2/'a;' = 1, y^x* = 1 and
, the like, approach the X- and
y-axes as asymptotes. The rates at which they approach the
^\
^=^^
^^m ^^
1
wMmm
~\
^
=^= — 1
Fig. 39. — A hyperbola
formed by capillary action of
two converging plane plates.
§25] THE POWER FUNCTION 61
axes depends upon the relative magnitudes of the exponents of the
powers of x and y; the quadrants in which the branches lie depend
upon the oddness or evenness of these exponents.
Exercises
1. Draw accurately upon squared paper the loci, xy = 1, xy' = 1,
x'y = 1. Use 2 centimeters as unit and make a pattern for xy = 1.
2. Show that the curves of the hyperboUc type lie in the rectangu-
lar regions shown plain, or not cross-hatched, in Fig. 35.
3. In what quadrants do the branches of x^^y'' = 1 he?
4. How does the locus of x^y^ — 1 differ from that oixy = 1 ?
6. Sketch, showing the essential character of each locus, the curves
x^y = 1, x^y = 1, s"""?/ = 1.
6. Show that xy = d passes through the point (\/a, -y/a) ', that xy
= a' passes through (a, a) and can be made from xy = Ihy "stretch-
ing" (if a > 1) both abscissas and ordinates of xy = 1 in the ratio
l:a.i
25. Symmetry. Some of the facts of symmetry respecting two
portions of the same parabola or hyperbola may be readily ex-
tended by the student to other curves. First answer the following
questions :
How are the points (a, 6) and ( — a, b) related to the F-axis?
How are the points (a, 6) and (a, —6) related to the X-axis?
How are the points (a, b) and (6, a) related to the line y = xl
Prove the result by plane geometry.
The following may then be readily proved by the student:
Theorems on Loci
I. 7/ X he replaced by (—x) in any eqvution containing x and y,
the new graph is the reflection of the former in the axis YY'.
II. If y be replaced by (—y) in any equation containing x and y,
the new graph is the reflection of the former in the axis XX'.
III. If x and y be interchanged in any equation containing x
and y, the new graph is the reflection of the former in the line y = x.
IV. If an equation remains unchanged when x is replaced by
{—x), its graph is symmetrical with respect to the Y-axis.
1 To "elongate" or "stretch" in the .ratio 2 :3 naeans to change tfie length of a
line segment so that (original length) : (new or stretched length) = 2:3.
62
ELEMENTARY MATHEMATICAL ANALYSIS [§26
v. If an equation remains unchanged when y is replaced by {—y),
its graph is symmetrical with respect to the X-axis.
VI. If an equation remains unchanged when x and y are inter-
changed, its graph is symmetrical with respect to the line y = x.
VII. If an equation remains unchanged when x is replaced by
i—x) and y by {—y), its graph is symmetrical with respect to the
origin.
VIII. If an equation remains unchanged when x is replaced by
(—J/) and y is replaced by (.—x), its graph is symmetrical with re-
spect to the line y = —x.
26. The Variation of the Power Function. The symmetry of
the graphs of the power function with respect to certain lines and
points, while of interest geometrically, nevertheless does not con-
stitute the most important fact in connection with these functions.
Of more importance is the law of change of value or the law by which
the function varies. Thus returning to a table of values for the
power function x^ for the first quadrant.
X
xi
d
0
0
1
1
1
4
?
4,
a
1
\
4
5.
3
g
4
1
%
7
2
¥-
5
¥
I
3
¥
i
-V-
5 .
4
¥-
i
we note that as x changes from 0 to ^ the function grows by the
small amount \. As x changes from \ by another increment of
\ to the value 1, the function increases by f to the value 1.
As X grows by successive steps or increments of'| unit each, it
is seen that x"^ grows by increasingly greater and greater steps,
until finally the change in x"^ produced by a small change in x
becomes very large. Thus the step by step increase in the func-
tion is a rapidly augmenting one, as is shown by the column of
§27] THE POWER FUNCTION 63
differences headed "d" in the table. Even more rapidly does the
function x^ gain in value as x grows in value. On the contrary,
for positive values of x the power functions 1/x, 1/x^, 1/x^, etc.,
decrease in value as x grows in value. Referring to the definition
of the slope of a curve given in §15, we see that the parabolic
curves have a positive slope in the first quadrant, while the hyper-
bolic curves have always a negative slope in the first quadrant.
The law of the power function is stated in more definite terms
in §34. That section may be read at once, and then studied
again in connection with the practical work which precedes it.
27. Increasing and Decreasing Functions. — As a point passes
from left to right along the X-axis, x increases algebraically.
As a point moves up on the F-axis, y increases algebraically and
as it moves down on the F-axis, y decreases algebraically. An
increasing function of x is one such that as x increases algebraic-
ally, y, or the function, also increases algebraically. By a
decreasing function of x is meant one such that as x increases
algebraically, y decreases algebraically. Graphically, an increas-
ing function is indicated by a rising curve as a point moves along
it from left to right. The power function y = s^ {n positive) is an
increasing function of x in the first quadrant and y = x~^ (— n
negative) is a decreasing function of x in the first quadrant.
The power function y = x^ \& an increasing function for all
positive atid for all negative values of x, while y = x'^\&& decreasing
function in the second quadrant but an increasing function in the
first quadrant. In a case like y = +xi, where y has two values
for each positive value of x, it is seen that one of these values
increases with x while the other decreases with x.
Exercises
1. Consider the function y = +x^. As x grows by successive steps
of one unit each, does the function grow by increasingly greater and
greater steps or not? Is the slope of the curve an increasing or a
decreasing function of x?
2. Does the algebraic value of the slope oi xy = 1 increase with x in
the first quadrant?
3. As £ changes from —5 to +5 does the slope oi y = x^ always
increase algebraically?
64 ELEMENTARY MATHEMATICAL ANALYSIS [§28
4. Express in the language of mathematics the fact that the curves
y = I", when n is a rational number greater than unity, are concave
upward.
Answer: "When n is greater than unity, the slope of the curve
increases as x increases."
Express in a similar way the fact that the curves y = x^/" are
concave downward.
28. The Graph of the Power Function when x» has a Coeffi-
cient. If numerical tables be prepared for the equations
y = x^
y' = 2x'
and
0\ 1 X> 2 -4 -3 -2 -1 O
(a) (b)
Fig. 40. — (o) The curves y = x' and y' = 2x'. (b) The curves y =
x^ and 2/ = (I) •
then for like values of x each ordinate of the second curve will
be two times the corresponding ordinate of the first curve. These
curves are shown in Fig. 40a. For each position of P on the
curve y' = 2x\ DP = 2DQ.
It is obvious that the curve
J i-u V = X" (1)
and the curve y, ^ a^„ ^2)
are similarly related; the ordinate of any point of the second graph
can be made from the corresponding ordinate (i.e., the ordinate
having the same abscissa) of the first graph by multiplying the
former by a. If o be positive and greater than unity, this corre-
§28] THE POWER FUNCTION 65
sponds to stretching or elongating all ordinates of (1) in the ratio
1 : o; if a be positive and less than unity, it corresponds to con-
tracting or shortening all ordinates of (1) in the ratio 1 : a.
For example, the graph of y' = ax' can be made from the
graph of y = x"ii the latter be first drawn upon sheet rubber, and if
then the sheet be uniformly stretched in the y direction in the ratio
1 : a. If the curve be drawn upon sheet rubber which is already
under tension in the y direction and if the rubber be allowed to
contract in the y direction, the resulting curve has the equation
y = ax" where a is a proper fraction or a positive number less than
unity.
The above results are best kept in mind when expressed in a
slightly different form. The equation y' = a-x" can, of course, be
written in the form (y'/a) = a;". Comparing this with the equa-
tion y = x", we note that (y'/a) = y or y' = ay, therefore we may
conclude generally that substituting (y'/a) for y in the equation of
any curve multiplies all of the ordinates of the curve by a. For
example, after substituting (y'/2) for y in any equation, the new
ordinate y' must be twice as large as the old ordinate y, in order
that the equation remain true for the same value of x.
(x'\ **
— I ,
that is, substituting ( — ) for x in any equation multiplies all of the
abscissas of the curve by a. See Fig. 406. Multiplying all
of the abscissas of a curve by a elongates or stretches all of the
abscissas in the ratio ^ 1 : a if a > 1, but contracts or shortens all
of the abscissas if o < 1. As the above reasoning is true for the
equation of any locus, we may state the results more generally
as follows:
Theorems on Loci
IX. Substituting ( -) for z in the equation of any locus multi-
plies all of the abscissas of the curve by a.
X. Substituting I - ) for y in the equation of any locus multiplies all
of the ordinates of the curve by a.
1 See footnote, p. 61.
66 ELEMENTARY MATHEMATICAL ANALYSIS [§29
Note: It is not necessary to retain the symbols x' and y' to
indicate new variables, if the change in the variable be otherwise
understood.
Exercises
1. Without actual construction, compare the graphs y, = a;^ and
^2 1 2
y = 5x2; J/ = a;2 and ^ = "2 ; 2/ = - and J/ = -; y = *» and y = 2x^;
s
y = x^ and 2/ = -g '
2. Without actual construction, compare the graphs y = x' and
2/ = f|V; 2/ = s' and ^= x^;y = x^a.ndy = {^j; y = x^ and | = x\
3. Compare y^ = a;= and i/^ = \k] ; j/^ = i' and (gj = x\- y' =x'
4. Compare the curves t/ = 2x^ and 2/ = 2 ( s j ; 3?/'' = a;' and
32/== (l) '; 2/^ = X' and (2)/)^' = {Zxy.
5. Compare ?/ = a; + 3 and y = 2 {x + 3); y = 2x — 1 and
I =2a; - 1; 2/ = 2a; - 1 and 22/ = 2a; - 1.
The following exercise involves a different principle from that used
above, which the student should reason out for himself.
6. Without actually constructing the curves, compare the curves,
for 2/ = 2a; + 3 and y = 2x + 5; y = x^ and y = x^ + l; y == x' and
2/ = a;' + 2; 2/ = a;' and y = x' — 1; y = x' and y = x' + i;
y =~ and y = — \- 2;y = x^ and 2/ = (x — 1)^.
29. Change of Unit. To produce the graph of 2/ = lOa;^ from
that of 2/ =^ a;^, the stretching of the ordinates in the ratio 1 : 10
need not actually be performed. If the unit of the vertical scale
of 2/ = a;^ be taken 1/10 of that of the horizontal scale, and the
proper numerical values be placed upon the divisions of the
scales, then obviously the graph ol y = x^ may be used for the
graph of 2/ = lOx''. Suitable change in the unit of measure on one
or both of the scales of 2/ = a?" is often a very desirable method of
representing the more general curve y — ax^.
An interesting example is given in Fig. 41. The period of vi-
THE POWER FUNCTION
67
1.6
1.4
■1.2
^
"^
-1.0
|0.8
|o.6
0.4
0 2
^
^
y
y
/
/
/
/
0 20 40 60 80 100 120 140 160 180 200
liength in Ooi.
Fig. 41.^ — Relation of length of a,
simple pendulum to period of vibration.
bration of a simple pendulum is given by the formula T = Tr-\/l/g.
When g' = 981 cm. per second per second (abbreviated cm. /sec. 2)
this gives T = O.W03-\/l, which for many purposes is sufficiently
accurate when written T = 0.10\/l. In this equation T must
be in seconds and I in centimeters. Thus when I = 100 cm., T
= 1 sec, so that the graph may be made by drawing the parabola
y = ^/x from the. pattern
previously made and then
attaching the proper num-
bers to the scales, as shown
in Fig. 41.
30. Variation. The re-
lation between y and a; ex-
pressed by the equation
y = ax", where n is any
positive number, is often
expressed by the state-
ment "y varies as the nth
power of X," or by the
statement "y is proportional to X"." Likewise, the relation
y = a/x", where n is positive, is expressed by the statement
"y varies inversely as the nth power of x." The statement "the
elongation of a coil spring is proportional to the weight of the
suspended mass" tells us
y = mx (1)
where y is the elongation (or increase in length from the natural
or unloaded length) of the spring, and x is the weight suspended by
the spring, hut it does not give us the value of m. The value of m
may readily be determined if the elongation corresponding to a
given weight be given. Thus if a weight of 10 pounds when sus-
pended from the spring produces an elongation of 2 inches in the
length of the coil, then, substituting x = 10 and y = 2 m (1),
and hence 2 = mlO
m = .
If this spring be used in the construction of a spring balance, the
length of a division of the uniform scale corresponding to 1 pound
will be 1/5 inch.
68 ELEMENTARY MATHEMATICAL ANALYSIS [§30
A special symbol, « , is often used to express variation. Thus
states that y varies inversely as d^. It is equally well expressed by
k
where A; is a constant called the proportionality factor.
The statements "y varies jointly as u and v," and "y varies
directly as u and inversely as v,'' mean, respectively,
y = kuv
hu
Thus the area of a rectangle varies jointly as its length and
breadth, or
A = kLB.
If the length and breadth are measured in feet and A in square feet,
k is unity. But, if L and B are measured in feet and A in acres,
then k = 1/43560. If L and B are measured in rods and A in
acres, then k = 1/160.
From Ohm's law, we say that the electric current in a circuit
varies directly as the electromotive force and inversely as the
resistance, or
C a -51 or C = A; ^•
K K
The constant multiplier is unity if C be measured in amperes, E
in volts, and R in ohms, so that for these units
^ - R
Exercises
1. The original length of a spring is 10 inches. The force, F,
necessary to stretch the spring is directly proportional to its elongation,
s. (o) Find the proportionality factor if a force of 200 pounds will
hold the spring at a length of 12 inches, (b) What force will be
required to hold the spring at a length of 13 inches? (c) What force
§31] THE POWER FUNCTION 69
will be required to elongate the spring 1 inch? Note that the elon-
gation is the extension of the length beyond the original length and not
the total length after elongation.
StfGGBSTioN: Since the force F is directly proportional to the elon-
gations, we may write
F = ks,
where k is the proportionality factor. We have given that F is 200
pounds when s is 2 inches.
2. Hooke's Law states that the elongation of a steel bar is propor-
tional to the force applied. A bar 500 inches long is stretched to a
length of 500.5 inches when a force of 1000 pounds is applied. Find
the proportionality factor.
3. Boyle's Law states that the volume of a perfect gas at constant
temperature varies inversely as the pressure. If volume is measured
as cubic feet and pressure as pounds per square inch, find the pro-
portionality factor if the volume is 13 cubic feet when the pressure
is 60 pounds per square inch. What will be the volume of the same
gas, according to Boyle's Law, if the pressure becomes only 15 pounds
per square inch?
31. Illustrations from Science. Some of the most important
laws of natural science are expressed by means of the power func-
tion' or graphically by means of loci of the parabolic or hyperbolic
type.
The linear equation y = mx is, of course, the simplest case of the
power function and its graph, the straight line, may be regarded as
the simplest of the curves of the parabolic type. The following
illustrations will make clear the importance of the power function
in expressing numerous laws of natural phenomena. Later the
student will learn of two additional types of fundamental laws of
science expressible by two functions entirely different from the
power function now being discussed.
The instructor will ask oral questions concerning each of the
following illustrations. The student should have in mind the
general form of the graph in each case, but should remember that
the law of variation, or the law of change of value which the func-
tional relation expresses, is the matter of fundamental importance.
The graph is useful primarily because it aids to form a mental pic-
ture of the law of variation of the function. The practical graph-
1 For brevity ax" as well as a:" will frequently be called a power function of x.
70 ELEMENTARY MATHEMATICAL ANALYSIS [§31
ing of the concrete illustrations given below will not be done at
present, but will be taken up later in §33.
(a) The pressure of a fluid in a vessel may be expressed in either
pounds per square inch or in terms of the height of a column of mer-
cury possessing the same static pressure. Thus we may write
p = 0.492/!, (1)
in which p is pressure in pounds per square inch and h is the height of
the column of mercury in inches. The graph' is the straight line
through the origin of slope 492/1000. The constant 0.492 can be
computed from the data that the weight of mercury is 13.6 times
that of an equal volume of water and that 1 cubic foot of water
weighs 62.5 pounds.
In this and the following equations, it must be remembered that
each letter represents a number, and that no equation can be used until
all the magnitudes involved are expressed in terms of the particular
units which are specified in connection with that equation.
(6) The velocity of a falling body which has fallen from a state of
rest during the time t, is given by
V = 32.2/, (2)
in which t is the time in seconds and v is the velocity in feet per second.
If t is measured in seconds and v is in centimeters per second, the
equation becomes' v = 98K. In either case the graph is a straight
line, but the lines have different slopes.
^ A full discuBsion of the process of changing formulas like the ones in the present
section into a new set of units should be sought in text-books on physics and mechan-
ics. The following method is sufficient for elementary purposes. First, write (for
the present example) the formula v = 32. 2£ where v is in ft./sec. and t is in seconds.
For any units of measure that may be used, there holds a general relation u = ct,
where c is a constant. To determine what we may call the dimensions of c, sub-
stitute for all letters in the formula the names of the units in which they are ex-
pressed, treating the names as though they were algebraic numbers. From v = ct
write, ft./sec, = csec. Hence (solving for dimensions of c), c has dimensions ft./sec.^
Therefore in the given case, we know c = 32.2 ft. /sec. 2. To change to any other
units simply substitute equals for equals. Thus 1 ft. = 30.5 cm., hence c = 32.2 X
30.5 cm./sec.2 = 981 cm./sec.^
To change velocity from mi./hr. to ft./sec. in formula (19) below, we have R =
0.00372 where R is in Ib./sq. ft. and V is in mi./hr. Write the general formula
R = cY^. The dimensions of c are (lb./ft.2) -7- (mi.Vhr.2) or (lb./ft.2) X (hr.^/mi.sy.
In the given case we have the value of c = 0.003 (lb. /ft. 2) X (hr.2/mi.2). To change
V to ft./sec, substitute equals for equals, namely 1 hr. = 3600 sec, 1 mi. = 5280
ft., which gives the formula R = 0.0013972^ where V is expressed as ft./sec and
R ig expressed as lb./ft.2. Note that 1 mi./hr. = f ft./sec. approximately.
§31] THE POWER FUNCTION 71
(c) The space traversed by a falling body is given by
s = igt\ (3)
or in English itaits (s in feet and t in seconds)
s = 16.1(2. ' (4)
(d) The velocity of the f aUing body, from the height h, is
V = ■\/2gh = V&iAh. (5)
The resistance of the air is not taken into account in formulas (2)
to (5).
The formula equivalent to (5)
jTOD^ = mgh, (6)
where to is the mass of the body, expresses the equivalence of ^mv^,
the kinetic energy of the body, and mgh, the work done by the force of
gravity mg, working through the distance h.
(e) The intensity of the attraction exerted on a unit mass by the
sun or by any planet varies inversely as the square of the distance
from the center of mass of the attracting body. If r stand for that
distance and if / be the force exerted on unit mass of the attracted
body, then
/ = ^- (7)
The constant m is the value of the force when r is unity.
(f) The formula for the horse power transmissible by cold-roUed
shafting is
where H is the horse power transmitted, d the diameter of the shaft in
inches, and N the number of revolutions per minute.
The rapid increase of this function (as the cube of the diameter)
accounts for some interesting facts. Thus doubling the size of the
shaft operating at a given speed increases 8-fold the amount of power
that can be transmitted, while the weight of the shaft is increased but
4-fold.
If H be constant, N varies inversely as d^ Thus an old-fashioned
5.0-h.p. overshot water-wheel making three revolutions per minute
requires about a 9-inch shaft, while a DeLaval 50-h.p. steam turbine
making 16,000 revolutions per minute requires a turbine shaft but
little over J^ inch in diameter.
72 ELEMENTARY MATHEMATICAL ANALYSIS [§31
(g) The period of the simple pendulum is
T = irVrTg, (9)
where T is the time of one swing in seconds, I the length of the pendu-
lum in feet, and g = 32.2 ft. /sec.', approximately.
(h) The centripetal force on a particle of weight W pounds, rotat-
ing in a circle of radius R feet, at the rate of JV revolutions per second is
F = ^ ^"^ , (10)
9
or if ff = 32.16 ft. /sec.?,
F = 1.227GWRN' (11)
where F is measured in pounds. If N be the number of revolutions
per minute, then
^ - 36009 ^^^^
= 0.000341 TFiJi\r2_ (13)
(i) An approximation formula for the indicated horse power required/
for a steamboat is
I.H.P. = ^, (14)
where S is speed in knots, D is displacement in tons, and C is a con-
stant appropriate to the size and model of the ship to which it is
appUed. The constant ranges in value from about 240, for finely
shaped boats, to 200, for fairly shaped boats.
(j) Boyle's law for the expansion of a gas maintained at constant
temperature is
pv = C, (15)
where p is the pressure and v the volume of the gas, and C is a constant.
Since the density of a gas is inversely proportional to its volume, the
above equation may be written in the form
P = cp, (16)
in which p is the density of the gas.
(fc) The flow of water over a trapezoidal weir is given by
q= Z.S7Lh^, (17)
where q is the quantity in cubic feet per second, L is the length of the
weir' in feet, and h is the head of water on the weir, in feet.
I The instructor is expected to explain the meaning of the terms here used.
§31] THE POWEE FUNCTION 73
{I) The physical law holding tor the adiabatic expanBion of air,
that is, the law of expansion holding when the change of volume is not
accompanied by a gain or loss of heat/ is expressed by
p = cp'-^'' (18)
This is a good illustration of a power function with fractional expo-
nent. The graph is not greatly different from the semi-cubical
parabola
y = ci'
(to) The pressure or resistance of the air upon a flat surface per-
pendicular to the current is given by the formula
R = 0.003F^ (19)
in which V is the velocity of the air in miles per hour and B is the
resulting pressure upon the surface in pounds per square foot. Ac-
cording to this law, a 20-mile wind would cause a pressure of about 1.2
pounds per square foot upon the flat surface of. a building. One foot
per second is equivalent to about 2/3 mile per hour, so that the formula
when the velocity is given in feet per second becomes :
R = 0.0013F2. (20)
(n) The power used to drive an aeroplane may be, divided into two
portions. One portion is utilized in overcoming the resistance of the
air to the onward motion. The other part is used to sustain the
aeroplane against the force of gravity. The first portion does "use-
less" work — ^work that should be made as small as possible by the
shapes and sizes of the various parts of the machine. The second part
of the power is used to form continuously anew the wave of compressed
air upon which the aeroplane rides. Calling the total power'' P, the
power required to overcome the resistance Pr, and that used to sus-
tain the aeroplane P«, we have
P =Pr+P, (21)
We learn from the theory of the aeroplane that P, varies as the cube
of the velocity, while P, varies inversely as V, so that
Pr = cV^ (22)
^ Note that when a vessel containing a gas is insulated by a non-conductor of
heat, so that no heat can enter or escape from the vessel, that the temperature of the
gaa will rise when the gas is compressed, and fall when it is expanded. Adiabatic
expansion may be thought of, therefore, as taking place in an inaulated vessel.
2 Power (= work done per unit time) is measured by the unit horse power, which
ia 550 foot-pounds per second.
74 ELEMENTARY MATHEMATICAL ANALYSIS [§31
and
P, =
k
(23)
Thus at high velocity less and less power is requireii to sustain the
aeroplane but more and more is required to overcome the frictional
resistance of the medium. The law expressed by (23) that less and
less power is required to sustain the aeroplane as the speed is increased
is known as Langley's Law. From this law Langley was convinced
25
24
.23
S22
S21
S20
§19
h18
|17
gl6
3 15
014
1 13
^112
Sll
Sio
39
g 8
« 6
■g 5-
fl 4
a
Hi 2
1
in 1 T iL_r rr r /^
M- t '^t t T _/J^z
3L txi t^^ 4 J Z^^/
W^Tf/ '! /b/ /-7/ /
^l^LJ ti Ij^Xj^ ^^
4ZJ-t tJ L/t/y^V
4 IJjty r /.ryY/^
J ^1^4 Tl_/^Tl/Zy&
ti^ tt-i/tj/L^ui^TO
ttl--4J^'tij.^\ftiy^4^
ttttj^-/^tKttZZC^
'^trr'^ttZt^Z-^t^
^ XlTH-JltZC^"^^^^
H 37^^555277^^53?^
Iir77Z55/:522^;^2^
tinmlt2z4>^t^
[I/QZ6Z232|g?^
IDAZgggglpJ^
ffizzpppp^
Wl /Aw^, ^^'
m™ ^^^^^
J aj^w
L
t
)Tl*lO«DC-000>o,-IOaCQ-3'»rt(ot-d60»^
Gals.for One Foot Depth
Fig. 42. — Capacity of rectangular and circular tanks per foot of
depth.
that artificial flight was possible, for the whole matter seemed to
depend primarily upon getting up sufficient speed. It is really this
law that makes the aeroplane possible. An analogous case is the
well-known fact that the faster a person skates, the thinner the ice
necessary to sustain the skater. In this case part of the energy of
the skater is continually forming anew on the thin ice the wave of
depression which sustains the skater, while the other part overcomes
the frictional resistance of the skates on the ice and the resistance of
the air.
§32] THE POWER FUNCTION 75
(o) The capacity of cast-iron pipe to transmit water is often given
by the formula
9'-88 = 1.68W-" (24)
in which q is the quantity of water discharged in cubic feet per second,
d is the diameter of the pipe in feet, and h is the loss of head measured
in feet of water per 1000 linear feet of pipe. This is a good illustra-
tion of the equation of a parabolic curve with complicated fractional
exponents. The curve very roughly approximates the locus of the
equation
y = cVhxi. (25)
(p) The contents in gallons of a rectangular tank per foot of depth,
6 feet wide and I feet long, is
q = 7.5W. (26)
The contents in gallons per foot of depth of "a cylindrical tank d feet in
diameter is
q = 7.5^^74. (27)
Fig. 42 shows the graph of (26) for various values of 6 and also shows
to the same scale the graph of (27).
32. Rational and Empirical Equations. — A number of the
formulas given above are capable of demonstration by means of
theoretical considerations only. Such for example are equations
(1), (2), (3), (4), (5), (7), (8), (9), (10), etc., although the constant
coefficients in many of these cases were experimentally deter-
mined. Formulas of this kind are known in mathematics as
rational equations. On the other hand certain of the above for-
mulas, especially equations (14), (17), (19), (22), (23), (24),
including not only the constant coefficients but also the law of
variation of the function itself, are known to be true only as the
result of experiment. Such equations are called empirical
equations. Such formulas arise in the attempt to express by an
equation the results of a series of laboratory measurements.
For example, the density of water (that is, the mass per cubic
centimeter or the weight per cubic foot) varies with the tem-
perature of the water. A large number of experimentors have
prepared accurate tables of the density of water for wide ranges
of temperature centigrade, and a number of very accurate empir-
ical formulas have been ingeniously devised to express the results,
of which the following four equations are samples :
76 ELEMENTARY MATHEMATICAL ANALYSIS [§33
Empirical fonnulas jor the density, d, of water in terms of tem-
perature centigrade, B.
96(9 - 4)2
(a) d = 1 -
10'
(K^ ^ 1 93(0 - 4)i»«^
(6) d = 1 jq^^
, , J , 6fl2 - 369 + 47
(c) "^ = 1 io«
, „ J , , 0.4859» - 81.39^ + 6029 - 1118
(d; d = 1 H jq^
Exercises
1. Among the power functions named in the above illustrations,
pick out examples of increasing functions and of decreasing functions.
2. Under the same difference of head or pressure, show by formula
(24) that an 8-inch pipe will transmit much more than double the
quantity of water per second that can be transmitted by a 4-inch pipe.
3. Wind velocities during exceptionally heavy hurricanes on the
Atlantic coast are sometimes over 140 miles per hour. Show that the
wind pressure on a flat surface during such a storm is about fifty times
the amount experienced during a 20-mile wind.
4. Show that for wind velocities of 10, 20, 40, 80, 160 miles per hour
(varying in geometrical progression with ratio 2), the pressure exerted
on a flat surface is 0.3, 1.2, 4.8, 19.2, 76.8 pounds per square foot
respectively (varying in geometrical progression with ratio 4) .
6. A 300-h.p. DeLaval turbine makes 10,000 revolutions per min-
ute. Find the necessary diameter of the propeller shaft.
6. A railroad switch target bent over by the wind during a tornado
in Minnesota indicated an air pressure due to a wind of 600 miles per
hour. Show that the equivalent pressure on a flat surface would be
7.5 pounds per square inch.
7. Show that a parachute 50 feet in diameter and weighing 50
pounds will sustain a man weighing 205 pounds when falling at the
rate of 10 feet per second.
Suggestion: Use approximate value ir = 22/7 in finding area of
parachute from formula for circle, nr^, and use formula (20) above.
8. Show that empirical formulas (a) and (6) for the density of
water reduce to a power function if the origin be taken at 9 = 4, d = 1.
33. Practical Graphs of Power Functions. The graphs of the
power function
2/ = a;^ y = x^, y = -> y = x\, etc., (1)
§33]
THE POWER FUNCTION
77
can, of course, be made the basis of the laws concretely expressed
by equations (1) to (27) of §31. If, however, the graph of a
scientific formula is to serve as a numerical table of the function
for actual use in practical work, then there is much more labor
in the proper construction of the graph than the mere plotting
of the abstract mathematical function. The size of the unit to
be selected, the range over which the graph should extend, the
permissible course of the curve, become matters of practical
importance.
If the apparent slope^ of
+ 1 or —1, it is desirable
to make an abrupt change
of unit in the vertical or
the horizontal scale, so as
to bring the curve back
to a desirable course, for
it is obvious that numeri-
cal readings can best be
taken from a curve when
it crosses the rulings of the
coordinate paper at ap-
parent slopes differing but
little from + 1.
The above suggestions
in practical graphing are
the follow-
a graph departs too widely from
E
350
■
i
'
1
\
/
'
/
/
(
'
>
1
/
'
>
/
/
/
ya
-A
7
ir
J
2 3 4 5 6 7 8
Diameter of Shaft in Inches
9 10
Pig. 43.— Capacitjr at 100 R.P.M.
of cold-rolled shafting to transmit
power.
illustrated by
ing example :
Graph the formula
(equation (8), §31), for the horse power transmissible by cold-
rolled shafting.
in which d is the diameter in inches and N is the number of
revolutions per minute. The formula is of interest only for the
range of d between 0 and 24 inches, as the dimensions of ordinary
1 Of course the real slope of a curve is independent of the scales used. By apparent
slope =3 1 is meant that the graph appears to cut the ruling of the squared paper
at about 45°.
78 ELEMENTARY MATHEMATICAL ANALYSIS [§33
shafting lie well within these limits. Likewise one would not
ordinarily be interested in values of N except those lying between
10 and 3000 revolutions per minute. Fig. 43 shows a suitable
graph of this formula for the range 1 < d < 10 for the fixed
value oiN = 100. In order properly to graph this function, three
different scales have been used for the ordinate H, so that the
slope of the curve may not depart too widely from unity.
If similar graphs be drawn for N = 200, N = 300, N = 400,
etc., a set of parabolas is obtained from which the horse power
of shafting for various speeds of rotation as well as for various
diameters may be obtained at once. A set of curves systematic-
ally constructed in a manner similar to that just described, is often
called a family of curves. Fig. 42 shows a family of straight lines
expressing the capacity of rectangular tanks corresponding to
the various widths of the tanks.
Inasmuch as many of the fqrmulas of science are used only for
positive values of the variables, it is only necessary in these cases
to graph the function in the first quadrant. For such problems
the origin may be taken at the lower left corner of the coordinate
paper so that the entire sheet becomes available for the curve in
the first quadrant.
The illustrations of §31 are sufficient to make clear the impor-
tance in science of the functions now being discussed. The follow-
ing exercises give further practice in the useful application of the
properties of the functions.
Exercises
The graphs for the following problems are to be constructed upon
rectangular coordinate paper. The instructions are for centimeter
paper (form Ml) ruled into 20 X 25 cm. squares. On other paper use
J inch in place of 1 centimeter. In each case the units for abscissa
and for ordinates are to be so selected as best to exhibit the functions,
considering both the workable range of values of the variables and '
the suitable slope of the curves.
The student should read §12 a second time before proceeding with
the following exercises, giving especial care to instructions (4), (5), and
(6) of that section.
1. Classify the graphs of formulas (1) to (27), §31, as to parabolic
or hyperbolic type.
§33] THE POWER FUNCTION 79
2. Graph the formula v^ = 2gh, or v = y/lgh = 8.02hi, if h range
between. 1 and 100, the second and foot being the units of measure.
See formula (5), §31.
The following table of values is readily obtained :
h\ 1 5 10 20 30 40 50 60 70 80 90 100
v\ 8.02 17.9 25.3 35.8 43.9 50.7 56.7 62.1 67.1 71.7 76.0 80.2
Use 2 cm. = 10 feet as the horizontal unit for h, and 2 cm. = 10 feet
per second as the vertipal unit for v. The graph is then readily con-
structed without change of unit or other special expedient.^
3. Graph the formula q = 3.37Lhi fori = 1, and for h = 0, 0.1,
0.2, 0.3, 0.4, 0.5. See formula (17), §31. Use 4 cm. = 0.1 for
horizontal unit for h and 2 cm. =0.1 for vertical unit for q.
4. Draw a curve showing the indicated horse power of a ship I.H.P.
= S'Di/C for C = 200 if the displacement D = 8000 tons, and for
the range of speeds iS = 10 to S = 20 knots. See formula (14), §31.
For the vertical unit use 1 cm. — 1000 h.p. and for the horizontal
unit use 2 cm. = 1 knot. Call the lower left-hand corner of the paper
the point (S = 10, I.H.P. = 0).
5. From the formula expressing the centripetal force in pounds of a
rotating body,
F = 0.000341 ITiJiV^
draw a curve showing the total centripetal force sustained by a 36-inch
automobile tire weighing 25 pounds, for all speeds from 10 to 40 miles
per hour. See formula (13), §31.
Miles per hour must first be converted into revolutions per minute
by .dividing 5280 by the circumference of the tire and then dividing
the result by 60. This gives
1 mile per hour = 9J revolutions per minute
If V be the speed in miles per hour the formula for F becomes
F = 0.000341(1.5)25(9J)2F2 = l.llF^
For horizontal scale let 4 cm. = 10 miles per hour and for the vertical
scale let 1 cm: = 100 pounds.
6. Draw a curve from the formula / = m/r'^ showing the accelera-
tion of gravity due to the earth at all points between the surface of
the earth and a point 240,000 miles (the distance to the moon) from
the center, if / = 32.2 when radius of the earth = 4000 miles.
It is convenient in constructing this graph to take the radius of
the earth as unity, so that the graph will then bo required of / = 32.2/r^
80
ELEMENTARY MATHEMATICAL ANALYSIS [§34
from r ■« 1 to r = 60. In order to construct a suitable curve several
changes of units are desirable. See Kg. 44. One centimeter repre-
sents one radius (4000 miles) from r = 0 to r = 10, after which the
scale is reduced so that 1 mm. represents one radius. In the vertical
direction the scale is 4 cm. = 10 feet per second for 0 < r < 5, 4 cm. =
1 foot a second for 5 < r < 10, and 4 cm. =0.1 foot a second for
10 < r < 60. Even with these four changes of units just used the
first and third curves are somewhat steep. The student can readily
improve on the scheme of Fig. 44 by a better selection of units.
40
80
20
90
I
I
\
1
I
\
N>
\
\
S
'
V
^
\,
\
^,
--.
\
s
"^
^
—
g 43
o
I 10
1 2 3 4 6 6 7 8 9 10 20 SO 40 EO 60
Distance from Earth's. Oeuter, Earth's Eadiuscl
Fig. 44. — Gravitational acceleration at various distances from the
earth's center. The moon is distant approximately 60 earth's radii
from the center of the earth.
34. The Law of the Power Functions. Sufficient illustrations
have been given to show the fundamental character of the power
function as an expression of numerous laws of natural phenomena.
How may a functional dependence of this sort be expressed in
words? If a series of measurements are made in the laboratory,
so as to produce a numerical table of data covering certain phe-
nomena, how can it be determined whether or not a power func-
tion can be written down which will express.the law (that is, the
function) defined by the numerical table of laboratory results?
§34] THE POWER FUNCTION 81
The answers to these questions are readily given. Consider first
the law of the falling body
s = 16.1i^ (1)
Make a table of values for values oi t = 1, 2, 4, 8, 16 seconds, as
follows :
t
1
2
4
8
16
s
16.1
64.4
257.6
1030.4
4121.6
The values of t have been so selected that t increases by a fixed
multiple, that is, each value of t in the sequence is twice the pre-
ceding value. From the corresponding values of s it is observed
that s also increases by a fixed multiple, namely 4.
Similar conclusions obviously hold for any power function.
Take the general case
y = ax", (2)
where n is any exponent, positive, negative, integral or fractional.
Let X change from any value xi to a multiple value mxi and call
the corresponding values of y, yi and 2/2. Then we have
2/1 = axi", (3)
and
2/2 = o(wia;i)" = aiwxi". (4)
Divide the members of (4) by the members of (3) and we have
^ = m». (5)
2/1
That is, if a; in any power function change by the fixed multiple
m, then the value of y will change by a fixed multiple w. Thus
the law of the power function may be stated in words in either of
the two following forms :
In any power function of x, if x change by a fixed multiple, y will
change by a fixed multiple also.
In any power function of x, if x increase by a fixed percent,
the function will increase or decrease by a fixed percent also.
This test may readily be applied to laboratory data to determine
whether or not a power, function can be set up to represent as a
formula the data in hand. To apply this test, select at several
places in one column of the laboratory data, pairs of numbers
which change by a selected fixed percent, say 10 per cent, or 20
82 ELEMENTARY MATHEMATICAL ANALYSIS 1§35
percent, or any convenient percent. Then the corresponding pairs
of numbers in the other column of the table must also be related by
a fixed percent (of course, not in general the same as the first-
named percent), provided the functional relation is expressible by
means of a power function. If this test does not succeed, then
the function in hand is not a power function.
Since the fixed percent for the function is ot" if the fixed percent
for the variable be m, the possibility of determining n exists,
since the table of laboratory data must yield the numerical values
of both m and to"
36. Simple Modifications of the Parabolic and of the H]rperbolic
Types of Curves. In the study of the motion of objects it is
convenient to divide bodies into two classes: first, bodies which
retain their size and shape unaltered during the motion; second,
bodies which suffer change of size or shape or both during the
motion. The first class of bodies are called rigid bodies ; a mov-
ing stone, the reciprocating or rotating parts of a machine, are
illustrations. The second class of bodies are called elastic bodies ;
a piece of rubber during stretching, a spring during elongation or
contraction, a rope or wire while being coiled, the water flowing in
a set of pipes, are all illustrations of this class of bodies.
When a body changes size or shape the motion is called a
strain.
Bodies that preserve their size and shape unchanged may possess
motion of two simple types: (1) Rotation, in which all particles
of the body move in circles whose centers lie in a straight fine
called the axis of rotation, which line is perpendicular to the plane
of the circles, and (2) translation, in which every straight line of
the body remains fixed in direction.
We have already noted that the curve.
(1)
2/1 =
ax'
or
a
s".
can
be made from the
curve
y =
x"
(2)
by multiplying all the ordinates of (2) by a. The effect is either
§36] THE POWER FUNCTION 83
to elongate or to contract all of the ordinates, depending upon
whether a > 1 or a < 1, respectively. The substitution of (j/i/a)
for y has therefore produced a motion or strain in the curve
y = x". Likewise
»=(?)■ («
can be made from
2/ = X" (4)
by multiplying all of the abscissas of (4) by a. The effect is
either to stretch or to contract all of the abscissas, depending
upon whether a > 1, or a < 1, respectively.
In general, if a curve has the equation
V = fix), (5)
then
(6)
»=/(?)
is nlade from curve (5) by ' lengthening or stretching the XY-
plane uniformly in the x direction in the ratio 1 : a.
The statement just given is made on- the assumption that
a>l. If a<l then the above statements must be changed
by substituting shorten or contract for elongate or stretch.
The reasons for the above conclusions have been previously
stated : substituting (— ) everywhere in the place of x multiplies
all of the abscissas by a. That is, if ( — I = x, then xi = ax, so that
Xi is a-fold the old x.
We shall now explain how certain of the other motions men-
tioned above may be given to a locus by suitable substitution for
X and y.
36. Translation of Any Locus. If a table of values be prepared
for each of the equations
as follows :
y = x' (1)
2/ = (xi - 3)2 (2)
x 1
-2 -10 12 3 4]
y 1
1
4 1 0 1 4 9 16
-2 -10 12 3 4 5 6
Xi
2/1 25 16 9 4 1 0 1 4 9
84 ELEMENTARY MATHEMATICAL ANALYSIS [§36
and then if the graph of each be drawn, it will be seen that the
curves differ ordy in their location and not at all in shape or size.
The reason for this is obvious: If {xi — 3) be substituted for x
in any equation, then since {xi — 3) has been put equal to x, it
follows that a;i = a; + 3, or the new x, namely Xi, is greater
Fig. 45. — The curve y^ = (x — f)^ is the curve y' = x' translated
to the right :| units.
than the original x by the amount 3. This means that the new
longitude of each point of the locus after the substitution is greater
than the old longitude by the "fixed amount 3. Therefore the
new locus is the same as the original locus translated to the right
the distance 3.
§36] THE POWER FUNCTION 85
The same reasoning applies if {xi — a) be substituted for x,
and the amount of translation in this case is a. The same reason-
ing applies also to the general case y = f{x) and y = f{xi — a),
the latter curve being the same as the former, translated the dis-
tance a in the x direction.
As it is always easy to distinguish from the context the new x
from the old x, it is not necessary to use the symbol Xi, since the
old and new abscissas may both be represented by x. The
following theorems may then be stated:
Theorems on Loci
XI. If {x — a) be substituted for x throughout any equation, the
ecus is translated a distance a in the x direction.
XII. If {y — 6) be substituted for y in any equation, the locus is
translated the distance b in the y direction.
These statements are perfectly general: if the signs of a and
6 are negative, so that the substitutions for x and y are of the form
X + a' and y + b', respectively, then the translations are to
the left and down instead of to the right and up.
Sometimes the motion of translation may seem to be disguised
by the position of the terms a or b. Thus the locus y = 3x -\- 5
is the same as the locus y ~ Sx translated upward the distance 5,
for the first equation is really y — 5 = 3x, from which the conclu-
sion is obvious.
Exercises
The student is not required to draw the curves in exercises 1 to 8 below,
but is expected to make the comparisons by means of the theorems on loci
given above.
1. Compare the curves: (1) y- - 2x and y = 2(z — I); (2) y = x'
and 2/ = (x — 4)'; (3) y = x^ and y — Z = x^; (4) ?/ = x^ and
y = (x - 5)1; (5) y = 5x^ and y = 5{x + 3)2; (6) y = 2x' and
y = 2{x - fc)'; (7) y = 2x' and y = 2x' + k; (8) y + 7 = x' and
y = x' and y - 7 = x'; (9) Sy" = Sx' and 3(,y - 6)^ = 5(x - a)\
2. Compare the curves: (1) y = x^ and y = {x/2)^; (2) y = x^
and y = x^/8; (3) y — x' and y/2 = x'; (4) y = x^ and y = 2x'; (5)
2/2 = 3x« and {.y/bY = 3(x/7)3; (6) y^ = x' and y'^ = (3i)'; (7)
2/ = x* and y — ix^ (note: explain in two ways); (8) y = x^ and
2y = x' and y = 27x».
86 ELEMENTARY MATHEMATICAL ANALYSIS [§36
3. Translate the locus y = 2a:'; (1) 3 units to the right; (2) 4 units
down; (3) 5 units to the left.
4. Elongate three-fold in the x direction the loci: (1) y^ = x; (2)
3y = x^; (3) y^ = 2x>; (4) y =2x + 7.
6. The loci named in exercise 4 have their ordinates shortened in
the ratio 2:1; write their equations.
6. Show that y = — -r-r and y = r are hyperbolas.
X ~i~ 0 X o
7. Show that y = — -j-r^ is a hyperbola.
Note : Divide the numerator by the denominator, obtaining the
b b
equation y = 1 ^—r> oi y — 1 = —
8. Show that y =^
or
x+b'"'" ' - x+b
X + a
xH>'
a — b
y = i+:
x+b
is a hyperbola, namely the curve xy = a — b translated to a new
position.
« oi i 1 , N ^ ,i_s ^ + 3 , ^ 3x + 2 ,
9. Sketch: (a) y = ^-^7^; (b) y = ^qrj; (c) V = "Jipj"; ^""
id) y = — 3-0"' Sketch a curve from which each curve is obtained
by translation.
10. Show how the graph for t/ = x^ + 4:X + 5 may be obtained from
the graph for y = x".
Hint: 2/ = x^ + 4x + 5 = x^ + 4x + 4 + 1 = (x + 2)^ + 1, or
2/ — 1 = (x + 2)2. Thus, the graph for y = x' + 4:X + 5 may be
obtained by translating the graph for y = x^ one unit up and two units
to the left.
11. Sketch the curves for:
(a) 2/ = x2 + 4x + 4; (b) y = x^ + 6x + 10;
(c) 2/ = x2 + 2x - 3; (d) 2/ = 4x2 4. 4^; + j.
(e) y = 4x' + 2x - 1; (f) y = ^x - x";
(s) 2/ = 6x - x"; (h) 2/ = x^ + 3x - 1;
(i) y ^2x^ +Zx; (j) j/ = 3x - 2x2;
(fc) 2/^ = X + 1.
12. Which of the curves of exercise 11 pass through the origin?
13. Sketch:
(a) x2 +2/2 = 1; (6) x^ + j/^ = 4;
(c) x2 + (2/ - i;2 = 1; (d) (x - 1)2 + 2/2 = 4;
(e) (x + 1)2 + (y - 2)2 = 5; (/) x2 + 2x + 2/* = 3.
§37]
THE POWER FUNCTION
87
37. Shearing Motion. Let the dotted curve Pi'OPi, Fig. 46
be the graph of the semicubical parabola y = xi and OP2 the graph
of y "= X. The graph P'OP is constructed by taking its ordinate,
for any value of the abscissa,
equal to the (algebraic) sum of
the ordinates of the two given
curves. Thus, DP = DPi +
DPi and DP' = DPi' + DP,.
The equation of P'OP is y =
a;t + X, since DPi = xi and
DP2 = X.
Exercises
1. From the curves for y = x^
and y = ^x, sketch y = x^ -\- |.-c.
2. From the curves for y = x'^
and 2/ = — |x, sketch y = x'' —
^x.
3. From the curves for y = —
x^ and y = x, sketch y = x — x^.
4. From the curves for y = -
X
and y = X, sketch y = — \- x-
5. From the curves for y = -
X ■
and y = X, sketch y = x
38. General Case. Consider
the production of the curve
y = fix) + mx (1)'
from the curve
Fig
— The shear of y
the line y = x.
and the straight line
y' = m
(2)
(3)
Graphically, the curve (1) is seen to be formed by the addition
of the ordinates of the straight line y" = mx to the corresponding
ordinates of y' = f(x) . Thus, in Fig. 47, the graph of the func-
tion a;^ + a; is made by adding the corresponding ordinates of
88
ELEMENTARY MATHEMATICAL ANALYSIS [§38
y = x^ and y = x. Mechanically, this might be done by draw-
ing the curve on the edge of a pack of cards (see Fig. 48), and then
slipping the cards over each other uniform amounts. The change
of the shape of a body, or the strain of a body, here illustrated, is
called lamellar motion or shearing motion. It is a form of motion
of very great importance.
4
m
\
0
/ / / "
/ / /a
/
\
Jl
A
-3 -2
I
Vx
X
A
! 3
4
-•>.
\
/
/
-3
\
■4
Fig. 47. — The shear of the cubical parabola 2/ = a' in the line y =
X, and also in the fine y = — x.
We shall speak of the locus y = f{x) + mx as the shear of the
curve y = f(x) in the line y = mx.
Theorems on Loci
XIII. The addition of the term mx to the right side of y = f{x)
shears the locus y = f(x) in the line y = mx.
The locus
y — ax^ + TOcc + 6
is made from y = a;' by a combination of first, a uniform elongation
THE POWER FUNCTION
89
[a], second,, a shearing motion [m], and third, a translation [6].
Either motion may be changed in sense by changing the sign of
a, m, or 6, respectively.
The student may easily show that the effect of a shearing motion
upon the straight line y = mx + b is merely a rotation about
the fixed point (0, b). The line is really stretched in the direction
M
H
K
8
7
0
5
4
3
2
1
-
o
o
o
0
0
o
0
2
o
o
o
I
o
0
)"
1
2
3
4
5
6
7
3
-
1
2
Fig. 48. — Shearing motion illustrated by the slipping of the members
of pack of cards.
of its own length, but this does not change the shape of the line
nor does it change the line geometrically. A line segment (that
is, a hne of finite length) would be modified, however.
The parabola y = x^ is transformed under a shearing motion
in a most interesting way. For, after shear, y = x' becomes
y = x^ + 2mx, (4)
where, for convenience, the amount of the shearing motion is
90
ELEMENTARY MATHEMATICAL ANALYSIS [§38
represented by 2w instead of by m. Writing. this in the form
y = x^ + 2mx + m^ — m^,
or
2/ = (x + my — m^,
y + m'' = (x + mY, (5)
we see that (4) can be made from the parabola y = x'^ hy trans-
lating the curve to the left the amount m and down the amount m^.
(See Fig. 49.)
\
\
4
S
"11
<='7 /
"V /
\
\
•>.
\\
-^
fe*/'
■3
-2
T
0
-1
.
I
3
y^
-9.
.9.
-4
Fig. 49. — The shear oi y = x^ in the line y = 0 . 6x.
Shearing motion, therefore, rotates the straight line and trans-
lates the parabola. The effect on other curves is much more
complicated, as is seen from Figs. 46 to 48.
The parabola y = x^ after shear is identical in size and shape
with y = x^ -\- mx + b. Likewise, y = ax' -\- bx + c is a para-
bola differing only in position from y = ax'.
Exercises
1. Explain how the curve y = x^ -\- 2x may be made from the
curve y = x^. How can the curve y = 2x' + 3x be made from the
curve y = 2x'?
§39] THE POWER FUNCTION 91
2. Find the coordinates of the lowest point oi y — x^ — ix, that
is, put this equation in the form y — b = {x — a)^.
3. Compare the curves y = x' -\- 2x and y = x^ — 2x. (Do not
draw the curves.)
4. Explain how the curve y = 1/x + 2x may be formed from the
curve y = 1/x and oi y = 2x.
•
, 39. Rotation of a Locus. The only simple type of displace-
ment of a locus not yet considered is the rotation of the locus
about the origin 0. This will be taken up in the next chapter.
40. Roots of Functions. The roots, or zeros, of a function are
the values of the argument for which the corresponding value of
the function is zero. Thus, 2 and 3 are rgots of the function
x^ — 5x + 6, for substituting either number for x causes the
function to become zero. The roots of a;^ — a; — 6 are + 3 and
- 2. The roots of x^ - 6x^ + llx - 6 are 1, 2, 3.
The word root, used in this sense, has, of course, an entirely
different significance from the same word in "square root," "cube
root,'' etc. But the roots of the function x'' — 5x — 6 are also the
roots of the equation x' — 5x — Q = 0.
In the graph of the cubic function y = x' — x in Fig. 47, the
curve crosses the X-axis at a; = — 1, x = 0, and x = 1. These
are the values of x that make the function x^ — x zero, and are, of
course, the roots of the function a;' — x. No matter what the
function may be, it is obvious that the intercepts on the X-axis of
the curve y = f{x), as OA, OB, Fig. 47, must represent the roots
off(x).
Exercises
1. From the curve y = x^ sketch the curves j/ — 4 = x^; i/ = 4x^;
^y = x^; y = (x - 4)2.
„ ^, , x' . , X* , (x — 3)2
2. Sketch y = -i^;y = ^^ - z] V = -2 - ^'<y = 2
ill 1
3. Sketch the curves y = x^; y = x^; y = 2x^; y = (x — 2)^;
y -2 = {x - 2)* and y = (x - 3)*.
4. Sketch the curves y' = (x-3)'; (2/ - 2)2 = x', and (y - 2)^ =
(x - 3)^
5. Graph yi = x and y^ = x' and thence y = x + x'.
6. Find the X-intercepts for the following :
[a) y = x^ + 2x - 3; (b) y = x' - x; (c) y = 2x'' + x - 3.
92 ELEMENTARY MATHEMATICAL ANALYSIS [§41
7. Find tbe roots of the following functions:
(a) x^ -Qx + 8; (6) x' +x -2;
(c) x' - X - 6; (d) 2x' - 5s + 2;
(e) Qx" +x - 1; (/) x^ + x'- 2x.
41. Intersectioii of Loci. Any pair of values of x and y that
satisfies an equation containing x and y locates some point on
the graph of that equation. Consequently, any set of values of
X and y that satisfies both equations of a system of two equations
containing x and y, must locate some point common to the
graphs of the two equations. In other words, the coordinates
of a point of intersection of two graphs is a solution of the equa-
tions of the graphs considered as simultaneous equations.
To find the values of x and y that satisfy two equations, we
solve them as simultaneous equations. Hence, to find the points
of intersection of two loci we must solve the equations of the
two curves. There will be a pair of values or a solution for each
point of intersection.
Thus, the intersection of the lines y = 3x — 2 and y = x/2 + 3
is the point (2, 4) and a; = 2, ?/ = 4, is the solution of the simul-
taneous equations.
Exercises
Find the point or points of intersection of the following pairs of loci:
l.y — X — S and y = 2x + 1.
2. y = x^ and y — x.
3. y = 2x^ and y = ix + 1.
4. 2/ = a;' and y — 2x.
5. y = — X and x' + y' = 2Sr.
Miscellaneous Exercises
1. Find the slope, the y-intercept, and the X-intercept for the
following:
Co) y =2x -3; (b) y = x + 2; (c) 3y - 6x = 10.
2. Write the equations of the lines determined by the following
data:
(o) slope 2 F-intercept 5
(ft) slope —2 y-intercept 5
(c) slope 2 y-intercept —5
(d) slope —2 y-intercept —5
(e) slope —2 X-intercept 4
§41] THE, POWER FUNCTION 93
3. Doesg the line 3y — 2x = 1 pass through the point :
(a) (1, 1); (6) (2, 2); (c) (-2, -1); (d) (0, 0); (e) ^3, 4).
4. Find the equation of a straight line with slope 2 and passing
through the point (3, 2).
5. Write the equations of the lines determined by the following
data:
(a) slope 1 and passing through (1, 1).
(6) slope —1 and passing through ( — 1, 1).
(c) slope 2 and passing through (1, —3).
{d) slope —3 and passing through (—2, —1).
6. Write the equation of a line passing through (2, 1) and (3, —5).
7. Write the equations of the lines passing through the following
pairs of points:
(a) a, 1) and (2, 3); ^6) (3, -1) and (-2, 1);
(c) (2, -3) and (2, 1); {d) (1, -5) and (-2, -3).
(e) (0, 2) and (3, 0); (/) (0, 0) and (-3, 2).
8. Make two suitable graphs upon a single sheet of squared paper
from the following data giving the highest and lowest average closing
price of twenty-five leading stocks listed on the New York Stock Ex-
change for the years given in the table:
Year
Highest
Lowest
1913
94.56
79.58
1912
101.40
91.41
1911
101.76
86.29
1910
111.12
86.32
1909
112.76
93.24
1908
99.04
67.87
1907
109.88
65.04
1906
113.82
93.36
1905
109.05
90.87
1904
97.73
70.66
1903
98.16
68.41
1902
101.88
87.30
Should smooth curves be drawn through the points plotted from this
table?
9. Plot the data given in following table upon squared paper. Use
the same horizontal axis for all three curves. Put the temperature
curves above the discharge curve, using the same horizontal (time)
scale for both. Let 1 cm. on the vertical scale represent 0.1 second-
94
ELEMENTARY MATHEMATICAL ANALYSIS [§41
foot' discharge, and 10° temperature. Start the temperature scale
with 60°, and place the 60, 6 cm. above the horizontal scale. Start
the discharge scale with 3.4 placed on the horizontal scale.
Discharge op a Seepage Ditch
Time,
Aug. 24, 1905
Discharge of ditch,
seo.-ft.
Temp, of water,
\-F.
Temp, of air.
8 : 00 a. m.
3.72
65
9:00 a.m.
3.70
70
74
11 :00 a. m.
3.66
79
84
1:30 p.m.
3.52
83
85
2:15 p.m.
3.49
3:30 p.m.
3.52
84
90
5:30 p.m.
3.66
78
88
6:00 p.m.
3.73
8:00 p. m.
3.84
67
76
10. Plot data given in the following table.'' Plot j^ along the
horizontal axis using 1 cm. to represent one-tenth unit. Plot veloci-
ties along the vertical axis, using 1 cm. to represent two one-hun-
dredths of 1 foot per second. Sketch a smooth curve among the
Relation Between Velocity and Depth at a Point in the
Lower Mississippi
Depth at observed point = d; whole depth = D
d
D
Velocity, feet per
second
d
D
Velocity, feet per
second
0.0
3.195
0.5
3.228
0.1
3.230
0.6
3.181
0.2
3.253
0,7
3.127
0.3
3.261
0.8
3.059
0.4
3.252
0.9
2.976
i
1 Second-foot (or sec.-ft.)i when applied to the measurement of flow of water
means one cubic foot per second.
2 The velocity at any point of a moving stream is determined by a current meter
placed at that point.
§41] THE POWER FUNCTION ' 95
points. The curve may not pass through all the plotted points.'
Begin to number the vertical axis with 2.90. The true origin will be
far below the sheet of paper.
11. Write the equations of the following curves after translated,
two units to the right; three units to the left; five units up; one unit
down; two units to the left and one unit down:
(a) y = 2x\ (6) y = -Zx'; (c) y = x^; {d) y =-
X
1 1 3 2
W2/=^; U)y=~^; (,g) y = x^-, (h) y = x^.
Sketch each curve in its original and also in its translated position.
12. Write the equation of each curve of exercise 11 when re-
flected in the X-axis; in the K-axis; in the line y = x. Sketch each
curve before and after reflection.
13. Shear each curve of exercise 11 in the line y = ix;in the line
y = — I x; in the line y = x; in the line y = — x. Sketch each curve
in its original and sheared position.
14. Draw on a sheet of coordinate paper the lines z = 0, x = 1,
x = —1, y = 0, y = 1, y = —1. Shade the regions in which the
hyperbolic curves lie with vertical strokes; and those in which the
parabolic curves lie with horizontal strokes. Write down all that
the resulting figure tells you.
15. Consider the following: y = x^, y = x~', y = xi, xy = — 1,
y = —x^,y^ = X*, y' = x% xy = I, x^ = — y', x* = — y'. In which
equation is y an increasing function of x in the first quadrant? For
which does the slope of the curve increase in the first quadrant? For
which does the slope of the curve decrease in the first quadrant?
16. Which of the curves of exercise 11 pass through (0, 0)? Through
(1, 1)? Through (-1, -1)?
17. Find the vertex of the curve y = x^ — 24x -|- 150.
Note : The lowest point of the parabola y = x^ may be called the
vertex.
Suggestion : It is necessary to put the equation in the form y — b
. =(x — ay. This can be done as follows: Add and subtract 144 on
the right side of the equation, obtaining
2/ = x^ - 24x + 144 - 144 + 150,
1 This curve is called » vertical velocity curve. In practical - work, however,
velocities are plotted along the horizontal axis and depths along the vertical axis,
and down from the origin. Your drawing gives the usual form of plotting if it is
turned 90" in a clockwise direction. Vertical velocity curves are parabolic in shape,
with the axis of the parabola parallel to the surface of the water. .
96 ELEMENTARY MATHEMATICAL ANALYSIS [|41
3/ « (s - 12)» + 6,
or
y -Q = {x - 12)'.
Then this is the carve y = x^ translated 12 units to the right and 6
units up. - Since the vertex oi y = x^ is at the origin, the vertex of the
given curve must be at the point (12, 6).
18. Find the vertex of the parabola V = x' — 6x + 11.
19. Find the vertex of ^ = x' + 8x + 1.
20. Find the vertex oi 4 + y = x' — 7x.
21. Find the vertex of ^ = 9x' + 18a; + 1.
22. Translate y = 4a;' — 12a; + 2 so that the equation will have
tke form y = 4x'.
23. Does the line 2y == 5x — 1 pass through the point of intersec-
tion of y = 2a; + 1 and y — 3x — 2t
CHAPTER IV
THE CIRCLE AND THE CIRCULAR FUNCTIONS
42. Equation of the Circle. In rectangular coordinates the
abscissa x, and the ordinate %, of any point P (as OD and DP,
Fig. 50) form two sides of a right triangle whose hypotenuse
squared is a;^ + j/'. If the point P move in such manner that the
length of this hypotenuse remains ,
fixed, the point P describes a - '^
circle whose center is the origin
(Fig. 50). The equation of this
circle is therefore
x2 + y2 = aS (1) ^,
where a = OP, the radius of the
circle.
It is sometimes convenient to
write the equation of the circle,
solved for y, in the form
y = + Va^ - xK (2)
This gives, for each value of x, the two corresponding equal and
opposite ordinates.
To translate the circle of radius a so that its center shall be at the
point (h, k), it is merely necessary to write
(x - h)'' + (y - k)^ = a^. (3)
This is the general equation of any circle in the plane XY, for it
locates the center at any desired point, {h, k) , and provides for any
desired radius a.
Exercises
1. Write the equations of the circles with center at the origin having
radii 3, 4, 11, V2 respectively.
2. Write the equation of each circle described in exercise 1 in the
form y = + s/a^ — x^.
7 97
Fig. 50. — The circle.
98 ELEMENTARY MATHEMATICAL ANALYSIS [§44
3. Which of the following points Jie on the circle s^ + j/^ = 169:
(5, 12), (0, 13J, C-12, 5), (10, 8), i9, 9), C9, 10)?
4. Which of the following points lie inside and which lie outside of
the circle x^ + y' = 100: ^7, 7), (10, 0;, (7, 8), (6, 8), (-5, 9),
(-7, -8), (2, 3), (10, 5), (\/40, VSO), (vlg, 9)?
43. The Equation, x^ + y^ + 2gx + 2fy + c = o (1)
may be put in the form (3) §42. For it may be written
x^ + 2gx + g^ + y^ + 2fy+P = g^+f-c
or
(x + gy + {v+ f)' ^ (Vg'+P-'c y, (2)
which represents a circle of radius -v/jf" +P — c whose center is at
the point (.— g, — /). In case g^ + P — c <.0, the radical
becomes imaginary, and the locus is not a real circle; that is,
coordinates of no points in the plane XY satisfy the equation. If
the radical be zero, the locus is a single point.
44. Any equation of the second degree, in two variables, lacking
the term xy and having like coefficients in the terms x^ and y^, repre-
sents a circle, real, null or imaginary. The general equation of
the second degree in two variables may be written
ax" + by' + 2hxy + 2gx + 2fy + c = 0. (3)
For, when only two variables are present, there can be present three
terms of the second degree, two terms of the first degree, and one
term of the zeroth degree. When a = b and h = 0 this reduces
to the form of (1) above after dividing through by a.
Exercises
Find the centers and the radii of the circles given by the following
equations:
1. x' + y' = 25. Also determine which of the following points are
on this circle: (3, 4), (5, 5), (4, 3), (-3, -4;, (-3, 4), (5, 0), (2, V2i).
2. x' + 2/2 =16. 4. x^ + y' -36 = 0.
3. x'+y'- -i = 0. 6. X' +y'' +2x = 0.
B. y = ± -\/l69 — x'. Also find the slope of the diameter through
the point (5, 12). Find the slope of the tangent at (5, 12).
§45] THE CIRCLE AND THE CIRCULAR FUNCTIONS 99
7. 9 - a;2 - 2/2 = 0. 10. (x + a)' + (y - b)' = 50.
8. x^ +y' -Qy = 16. 11. x' + y^ + 6x - 2y ^ 10.
9. x^ -2x +y^ - &y = 16.. 12. x'' + y' - ix + 6y = 12.
13. x2 + j/2 - 4x - 82/ + 4 = 0.
14. 3x' + 3y' + 6x + 12y - 60 = 0,
16. Is x^ + 2y^ +3x — 4:y — 12 = Q the equation of a circle?
Why?
16. Is 2x' + 2y' — 3x + 4y — 8 = 0 the equation of a circle?
Why?
45. Angular Magnitude. By tHe magnitude of an angle is
meant the amount of rotation of a line about a fixed point. If
a line OA rotate in the plane XY about the fixed point 0 to the
Fig. 51. — Positive angles.
Fig. 52. — Negative angles.
position OP, the line OA is called the initial side and the hne OP is
called the terminal side of the angle AOP. The notion of angular
magnitude as introduced in this definition is more general than
that used in elementary geometry. There are two new and very
important consequences that follow therefrom:
(1) Angular magnitude is unlimited in respect to size — that is,
it may be of any amount whatsoever. An angular magnitude
of 100 right angles or twenty-five complete rotations is quite
as possible, under the present definition, as an angle of smaller
amount.
(2) Angular magnitude exists, under the definition, in two
opposite senses — for rotation may be clockwise or anti-clockwise.
100 ELEMENTARY MATHEMATICAL ANALYSIS [§46
As is usual in mathematics, the two opposite senses are distin-
guished by the terms positive and negative. If the rotation is
anti-clockwise the angle is positive; if clockwise it is negative.
In Fig. 51, AOPi, AOPi, AOP3, and AOPt are positive angles.
In designating an angle its ihitial side is always named first. Thus,
in Fig. 51, AOPi designates a positive angle of initial side OA.
In Fig. 52, AOPi, AOPi, AOP3, and AOPt are negative angles.
In Cartesian coordinates, OX is usually taken as the initial
line for the generation of angles. If the terminal side of an
angle falls within the first quadrant, the angle is said to be an
angle of the first quadrant. If the terminal side of any
angle falls within the second quadrant, it is said to be an "angle
of the second quadrant," etc.
Two angles which differ by any multiple of 360° are called
congruent angles. We shall find that in certain cases congruent
angles may be substituted for each other without modifying results.
The theorem in elementary geometry, that angles at the
center of a circle are proportional to the intercepted arcs, holds
obviously for the more general notion of angular magnitude here
introduced.
46. Units of Measure. Angular magnitude, like all other
magnitudes, must be measured by the application of a suitable
unit of measure. Four systems are in common use:
(1) Right Angle System. Here the unit of measure is the right
angle, and all angles are given by the number of right angles and
fraction of a right angle therein contained. This unit is famihar
to the student from elementary geometry. A practical illus-
tration is the scale of a mariner's compass, in which the right angles
are divided into halves, quarters and eighths.
(2) The Degree System. Here the unit is the angle corre-
sponding to xriT of a complete rotation. This system, with the
sexagesimal sub-division= (division by 60ths) • into minutes
and seconds, is familiar to the student. This system dates back
to remote antiquity. It was used by, if it did not originate
among, the Babylonians.
(3) The Hour System. In astronomy, the angular magnitude
about a point is divided into 24 hours, and these into minutes
THE CIRCLE AND THE CIRCULAR FUNCTIONS 101
and seconds. This system is analogous to our system of measuring
time.
(4) The Radian, or Circular System. Here the unit of measure
is an angle such that the length of the arc of a circle described about
the vertex as center is equal to the length of the radius of the
circle. This system of angular measure is fundamental in me-
chanics, mathematical physics and pure mathematics. It must
be thoroughly mastered by the student. The unit of measure in
this system is called the radian. Its size is shown in Fig. 53.
O Radius
Fig. 53. — Definition of the Radian. The angle AOP is one Radian.
Inasmuch as the radius is contained 2ir times in a circumference,
we have the equivalents:
2% radians = 360°.
or 1 radian = 57° 17' 44".8 = 57° 17'.7 = 57°.3 nearly.
1 degree = 0.01745 radians.
The following equivalents are of special importance:
a straight angle = x radians.
a right angle = ^ radians.
60° = ^ radians.
o
45°
radians.
30° = ^ radians,
o
102 ELEMENTARY MATHEMATICAL ANALYSIS [§47
There is no generally adopted scheme for writing angular magni-
tude in radian measure. We shall use the superior Roman letter
"'" to indicate the measure. For example, 18° = O.SMW.
Since the circumference of a circle is incommensurable with its
diameter, it follows that the number of radians in an angle is
always incommensurable with the number of degrees in the angle.
The speed, or angular velocity, of rotating parts is usually
given either in revolutions per minute (abbreviated "r.p.m.")
or in radians per second.
47. Uniform Circular Motion. Suppose the line OP, Fig. 50,
is revolving counter-clockwise at the rate of ¥ per second, the
angle AOP in radians is then ht, t being the time in seconds re-
quired for OP to turn from the initial position OA. If we call d
the angle AOP, we have B = kt&s the equation defim'ng the motion.
The following terms are in common use:
1. The angular velocity of the uniform circular motion 's k
(radians per second).
2. The amplitude of the uniform circular motion is a.
3. The period of the uniform circular motion is the number of
seconds required for one revolution.
4. The frequency of the uniform circular motion is the number
of revolutions per second.
Sometimes the unit of time is taken as one minute. Also the
motion is sometimes clockwise, or negative.
Exercises
1. Express each of the following in radians: 135°, 330°, 225°, 15°,
150°, 75°, 120°. (Do not work out in decimals; use jr).
2. Express each of the following in degrees: 0.2'', ^u^' •fTr''' ^ir'-
3. How many revolutions per minute is 20 radians per second?
4. The angular velocity, in radians per second, of a 36-inch auto-
mobile tire is required, when the car is making 20 miles'per hour.
6. What is the angular velocity in radians per second of a 6-foot
drive-wheel, when the speed of the locomotive is 50 miles per hour?
6. The frequency of a cream separator is 6800 r.p.m. What is
its period, and its angular velocity in radians per second?
7. A wheel is revolving uniformly 30^ per second. What is its per-
iod and frequency?
§48] THE CIRCLE AND THE CIRCULAR FUNCTIONS 103
8. The speed of the turbine wheel of a 5-h.p. DeLaval steam turbine
is 30,000 r.p.m. What is the angular velocity in radians per second?
48. The Circular, or Trigonometric Functions. To each point
on the circle x^ + y^ = a? there corresponds not only an abscissa
and an ordinate, but also an angle 6 < 360°, as shown in Fig. 50.
This angle is called the direction angle, or vectorial angle, of the
point P. When 9 is given, x, y, and a are not determined, but the
ratios y/a, x/a, y/x, and their reciprocals, a/y, a/x, x/y are de-
termined. Hence these ratios are, by definition(§6), fimctions
of d. They are known as the circular, or trigonometric, functions
of 6, and are named and written as follows:
Function of e
Name
Written
y/a.
sine of 0.
sin 0.
x/a.
cosine of 0.
cos 0.
y/x.
tangent of 0.
tan 0.
x/y.
cotangent of 0.
cots.
a/x.
secant of 0.
sec0.
a/y.
cosecant of 0.
CSC 0.
The circular functions are usually thought of in the above order;
that is, in such order that the first and last, the middle two, and
those intermediate to these, are reciprocals of each other.
The names of the six ratios must be committed to memory.
They should be committed, using the names of x, y, and a as
follows :
Ratio Written
ordinate/radius. sin 0.
abscissa/radius. cos 0.
ordinate/abscissa. tan 0.
abscissa/ordinate. ' cot 0.
radius/abscissa. sec 0.
radius/ordinate. esc 0.
The right triangle DOP, Fig. 50, of sides x, y, and a, whose ratios
give the functions of the angle XOP, is often called the triangle of
reference for this angle. It is obvious that the size of the triangle
of reference has no eifect of itself upon the value of the functions
of the angle. Thus in Fig. 51 either DiOPi or D/OPi' may be
104 ELEMENTARY MATHEMATICAL ANALYSIS
taken as the triangle of reference for the angle 6. Smce the
triangles are similar we have
P^D, P.'Di' P^D, Pi'D,'
ODi OD,'
OPi OPi'
etc., which shows that identical ratios or trigonometric functions of
6 are derived from the two triangles of reference.
49. 1 Elaborate means for computing the six functions have been
devised and the values of the functions have been placed in
convenient tables for use. The functions are usually printed to
3, 4, 5 or 6 decimal places, but tables of 8, 10 and even 14 places
exist. The functions of only a few angles can be computed by
o^V?
i'lG. 54. — Triangles of reference for angles of 30°, 45°, and 60°
elementary means; these angles are, however, especially important.
(1) The Functions of 30°. In Fig. 54a, if angle AOB be 30°,
angle ABO must be 60°. By constructing the equilateral triangle
BOB', each angle of triangle BOB' will be 60°, and
y = AB = ^BB' = ia.
Therefore '
sin 30°
Also,
OA = \0B^ - AB^ = Va^ - ia^ = iaVs.
Therefore
sin 30° = i,
1 Some will prefer to take §50 before §49.
§49] THE CIRCLE AND THE CIRCULAR FUNCTIONS 105
o 2 '
tan 30° = ^^- = ^,
1 30° ^ ^'
1 2V3
•=°*3«° = t^ = ^/3,
sec 30° =
cos 30° 3 '
CSC 30° = ^^ = 2.
sm 30
(2) Functions of 45°. In the diagram, Fig. 546, the triangle
AOB is isosceles, or y = a;, and a^ = x^ + y' = 2x^. It follows
that a = X •\/2 = y -\/2-
Therefore
sin45° = ^=^,
2/ \/2 2 \
cos 45°
a; V2
xV2 2 '
tan 45°
= '- = !,-
X
cot 45°
^ 1
tan 45° '
sec 45°
~ cos 45° ~ ^^'
CSC 45°
-sin 45° =^2.
(3) Functions of 60°. In the diagram. Fig. 54c, construct the
equiangular triangle B'OB; then it is seen that, as in case (1)
above,
OA = h OB' = h a-
and
y = VaT^^^Ya^ = i a\/3.
Therefore
sin 60° = *"^^ ^ V3
a 2
106 ELEMENTARY MATHEMATICAL ANALYSIS [§50
cos 60° = ^ = i
tan 60°-^''^ =
50
= V3.
cot 60° —
_ V3
tan 60°
3
r-nr Rf\°
= 2.
COS 60°
CSC 60° - ^
. 2V3
sin 60°
50. Graphical Computation of Circular Functions. Approxi-
mate determination of the numerical values of the circular func-
tions of any given angle may be made graphically on ordinary
coordinate paper. Locate the vertex of the angle at the inter-
section of any two lines of the squared paper, form Ml. Let
the initial side of the angle coincide with one of the rulings of the
squared paper and lay off the terminal side of the angle by means
of a protractor. If the sine or cosine is desired, describe a circle
about the vertex of the angle as center using a radius appro-
priate to the scale of the squared paper — for example, a radius of
10 cm. on coordinate paper ruled in centimeters and fifths (form
Ml) permits direct reading to j-^ of the radius a and, by interpo-
lation, to j^u" of the radius a. The ordinate and abscissa of the
point of intersection of the terminal side of the angle and the circle
may. then be read and the numerical value of sine and cosine com-
puted by dividing each of these by the length of the radius.
If the numerical value of the tangent or cotangent be required,
the construction of a circle is not necessary. The angle should
be laid off as above described, and a triangle of reference con-
structed. To avoid long division, the abscissa of the triangle of
reference may be taken equal to 50 or 100 mm. for the determina-
tion of the tangent; and the ordinate may be taken equal to 50
or 100 mm. for the determination of the cotangent.
The following table (Table III) contains the trigonometric
functions of acute angles for increments of 10° of the argument
from 0° to 90°. (See the end of the book for a larger table.)
§50] THE CIRCLE AND THE CIRCULAR FUNCTIONS 107
Table III
Natural Trigonometric Functions to Two Decimal Places
0'.
9'
sin d
cos 0
tan e
cot 0
sec 8
CSC d'
0
0.00
0.00
1.00
0.00
00
1.00
03
10
0.17
0.17
0.98
0.18
5.67
1.02
5.76
20
0.35
0.34
0.94
0.36
2.75
1.06
2.92
30
0.52
0.50
0.87
0.58
1.73
1.15
2.00
40
0.70
0.64
0.77
0.84
1.19
1.31
1.56
50
0.87
0.77
0.64
1.19
0.84
1.56
1.31
60
1.05
0.87
0.50
1.73
0.58
2.00
1.15
70
1.22
0.94
0.34
2.75
0.36
2.92
1.06
80
1.40
0.98
0.17
5.67
0.18
5.76
1.02
90
1.57
1.00
0.00
CO
0,00
OO
1.00
The most important of these results are placed in the following
table :
0°
30°
45°
60°
90°
Sine
0
1
2
V2
2
V3
2
1
Cosine
1
V3
2
V2
2
1
0
Tangent. .
0
V3
3
1
V3
CO
V2 =
1.4142
V3-
= 1.7321
Exercises
1. Find by graphical construction all the functions of 15°.
Note. — A protractor is not needed as angles of 45° and 30° may be
constructed with ruler and compass.
2. Find tan 60°. Compare with the value found above in § 49 .
3. Lay off angles of 10°, 20°, 30°, and 40° with a protractor and
determine graphically the sine of each angle; record the results in a
suitable table.
4. Find the sine, cosine, and tangent of 75°.
6. Which is greater, sec 40° or esc 60°?
6. Determine the angle whose tangent is \.
7. Find the angle whose sine is 0.6.
108 ELEMENTARY MATHEMATICAL ANALYSIS [§S1
8. Which is greater, sin 40° or 2 sin 20°?
9. Does an angle exist whose tangent is 1,000,000?
61. Signs of the Functions. The circular functions have, of
course, the algebraic signs of the ratios that define them. Of the
three numbers entering these ratios, the distance, or radius
a, is always to be taken as positive. It enters the ratios, there-
fore, always as a positive number. The abscissa and the ordinate,
X and y, have the algebraic signs appropriate to the quadrants in
which P falls. The student should determine the signs of the
functions in each quadrant, as follows: (See Fig. 50.)
First
quadrant
Second
quadrant
Third
quadrant
Fourth
quadrant
Sine
+
+
+
+
+
+
Cosine
Tangent
Of course the reciprocals have the same signs as the original
functions.
The signs are readily remembered by the following scheme:
Sine Cosine Tangent
+
+
+
+
+
Cosecant
Secant
+
Cotangent
52. Triangles of reference, geometrically similar to those in
Fig. 54 for angles of 30°, 45°, and 60°, exist in each of the four
quadrants, namely, when the hypotenuse of the triangle of
reference in these quadrants is either parallel or perpendicular
to the hypotenuse of the triangle in the first quadrant. Then
an acute angle of one must equal an acute angle of the other
and the triangles must be similar. The numerical values of the
functions in the two quadrants are therefore the same. The
algebraic signs are determined by properly taking account of
§53] THE CIRCLE AND THE CIRCULAR FUNCTIONS 109
the signs of the abscissa and the ordinate in that quadrant.
Thus the triangle of reference for 120° is geometrically similar
to that for 60°. Hence, sin 120° =
and tan 120° = - Vs.
Exercises
V3
, but cos 120° = - i,
1. The student wiU fill in the blanks in the following table with
the correct numerical value and the correct sign of each function:
Function
120°
135°
150°
210°
225°
240°
300°
315°
330°
Sin
Cos
Tan
Cot
Sec
Csc
2. Write down the functions of 390° and 405°.
3. The tangent of an angle is 1. What angle < 360° may it be?
4. Cose = — |. What two angles < 360° satisfy the equation?
5. Sec 0 = 2. Solve for all angles <360°.
6. ^Csc e = - y/% Solve for B < 360°.
53. Functions of 0° and 90°. In Fig. 50, let the angle AOP
decrease toward zero, the point P remaining on the circumference
of radius a. Then y or PD decreases toward zero. Therefore,
sin 0° = 0. Also x or OD increases toward the value a so that
the ratio x/a becomes unity, or cos 0° = 1. Likewise the ratio
y/x becomes zero, or tan 0° = 0.
The reciprocals of these functions change as follows: As the
angle AOP approaches zero, the ratio a/y increases in value
without limit, or the cosecant becomes infinite. In symbols
(see §24) csc 0° =^ w , Likewise) cot 0° =5 «> , but sec 0° = L
110 ELEMENTARY MATHEMATICAL ANALYSIS [§54
In a similar way the functions of 90° may be investigated. Tiie
following table gives the variation of the functions as the angle
varies from 0° to 90°, from 90° to 180°, etc.:
Angle
From
0° to 90°
From
90° to 180°
Prom
180° to 270°
From
270° to 360°
Sin
Cos
Tan
Cot
Sec
Oto + 1
+ Ito 0
0 to +«>
+ 0O to 0
+ 1 to + 00
+ CO to + 1
+ Ito 0
Oto - 1
-o= to 0
0 to - oo
— oo to — 1
+ 1 to + oo
Csc
The student is to supply the results for the last two coltimns.
54. Fundamental Selations. The trigonometric functions are
not independent of each other. Because of the relation x^ + j/^
= a'-, it is possible to compute the numerical or absolute values of
the remaining five functions when the value of any one of the six
is given. This may be accompHshed by means of the fundamental
formulas derived below.
Divide the members of the equation
by a? Then
or,
a:^ + 2/2 =
1,
sin" e + cos2 e = I.
Likewise divide (1) through by x^; then
' + ©"=©"
or
sec2 0 = 1+ tan2 B.
Also divide (1) through by y'^; then
cgc^ ? = 1 + cot= 9.
or
(1)
(2)
(3)
(4)
THE CIRCLE AND THE CIRCULAR FUNCTIONS 111
Also, since
we obtain
and likewise
y
a
X
a
tan e =
cot e =
y
sin 6
cos d
cos 6
(5)
(6)
Formulas (2) to (6) are the fundamental relations between the six
trigonometric functions. They must be committed to memory
by the student.
= tan csc-=l + cot-
FiG. 55. — Diagram of the relations between the six circular functions.
sin
cos
The above relations between the functions may be illustrated
by a diagram as in Fig. 55. The simpler, or reciprocal, relations
are shown by the connecting lines drawn above the functions.
The reciprocal equations and the formulas (2), (3), and (4) are
sufficient to express the absolute or numerical value of any function
of any angle in terms of any other function of that angle. The
algebraic sign to be given the result must be properly selected
in each case according to the quadrant in which the angle lies.
Exercises
All angles in the following exercises are supposed to be less than
ninety degrees.
112 ELEMENTARY MATHEMATICAL ANALYSIS [§54
1. Sin e = 1/5. Find the values for the other five circular
functions.
Draw a right triangle whose hypotenuse is 5, whose altitude is 1
and whose base coincides with OX. In other words, make a = 5
and y = 1 in Fig. 56. Calculate x = v'25 — 1 = 2 \/6 and write
down all of the functions from their definitions.
2. Cos 9 = 1/3. Find esc 9.
Take a = 3 and a; = 1 in Fig. 56. Find y and then write down the
function from its definition.
3. Tan 9 = 2. Find sin d.
Take x = I and y = 2 in Fig. 56, calculate a, and then write
down the function from its definition.
O X A
Fig. 56. — Triangle of reference for B and for complement of S.
4. Sec e = 10. Find esc 6.
Take a = 10 and x = 1 and compute y.
6. Find the values of all functions of 9 if cot 6 = 1.5.
6. Find the functions of 9 if cos 8 = 0.1.
7. Find the values of each of the remaining circular functions in
each of the following cases:
(a) sin e = 5/13. (d) tan e = 3/4.
(6) cos e = 4/5.
(e) sec 9 = 2.
(g) tan 6 = m.
{h) sin e =
Vc" + d'
(c) sec 0 = 1.25. (/) tan e = 1/3.
Show that the following equalities are correct:
8. tan d cos 9 = sin 6.
9. sin e cot B sec 9 = 1.
10. (sin 9 + cos 9)2 = 2 sin 9 cos 9 + 1.
11. tan 9 + cot 9 = sec 9 esc 9.
12. Express each trigonometric function in terms of each of the
others; i.e., fill in all blank spaces in the following table;
§54] THE CIRCLE AND THE CIRCULAR FUNCTIONS 113
sin
cos
tan
cot
sec
cso
sin
sin
1
esc
cos
cos
1
sec
tan
tan
1
cot
cot
1
tan
cot
sec
1
cos
sec
CSC
1
sin
CSC
Fig. 57. — Diagram for exercise 13.
The following exercises refer to angles <360° of any quadrant:
13. If sin 9 = — 3/4 and tan B is positive, find the remaining five
functions.
Hint: Since sin e is negative and tan 9 is positive, the angle 9 is in
the third quadrant. See Fig. 57.
8
114 EaiEMENTARY MATHEMATICAL ANALYSIS [§55
14. If C08 9 = 12/13 and sin 9 ia negative, find the remaining five
functions of 6.
15. If tan e = — \/3 and cos B is negative, find the remaining func-
tions of e.
16. If cos 9 = — 1/3 and sin 9 is positive, find the remaining
functions.
17. If tan 9 = 5/12 and sec 9 is negative, find the remaining
functions of 9.
18. If sin 9 = 3/5 and tan 9 is negative, find the jemaining func-
tions of 9.
Pl(k,h)
P lh,/c)
Fig. 58. — Triangles of reference for complementary angles.
65. Functions of Complementary Angles. Angles are said
to be complementary if their sum is 90°. Angles are said to be
supplementary if their sum is 180°.
Let 0 be an angle in the first quadrant, and draw the angle
(90° -0) of terminal side OPi, as shown in Fig. 58. Let P and Pi
lie on a circle of radius a. Let the coordinates of the point P be
{h, k), then Pi is the point {k, h). Hence PiDi/OPi = h/a =
sin (90° -5). But from the triangle PDO, h/a = cos 8. Hence
Likewise,
sin (90° — 6) = cos 6
tan (90° - 0) = cot e
sec (90° - d) = esc e
These relations explain the meaning of the words cosine, co-
tangent, cosecant, which are merely abbreviations for comple-
§56] THE CIRCLE AND THE CIRCULAR FUNCTIONS 115
merit's sine, complement's tangent, etc. Collectively, cosine,
cotangent, and cosecant are called the co-functions. Likewise,
from Fig. 58,
cos (90° — 6) = siad
cot (90° - 0) = tan e
CSC (90° — d) = sec d
Later it will be shown that the above relations hold for all
values of d, positive or negative.
56. Graph of the Sine and Cosine. In rectangular coordinates
we can think of the ordinate y of a point as depending for its
value upon the abscissa or x of that point by means of the equation
y = sin X, provided we think of each value of the abscissa laid
off on the Z-axis as standing for some amount of angular mag-
nitude. Therefore the equation y = sinx must possess a graph
Y
A
e
C
p
^
-
-
7
^
-
-
i.
--
--
-
-
-
-
-
-
-
-
-
-
-
-
-
A
/
s
^ \\
A
/
«
-TT
A
/
1/
s
p
0
s.
/v
B D,D/^
1
s
/
A
s
/
y
/
\
/
y
/
\
/
y y
^_
-
-
-
-
—
^
-
-
-
-
^
-
--
-q
|h
-
-
-
-
V
s
-
-
-
-
-
-^
2
^
-
-
=
X
L.
_
_
_
„
_
_
_
_
_
a
5
^^
2-1
r-
_
_
„
^
_
_
^
^
T-
^
_
_
^
->
Fig. 59. — Construction of the sinusoid.
in rectangular coordinates. In order to produce the graph of
y = sin X, it is best to lay off the angular measure x on the X-
axis in such a manner that it may conveniently be thought of
in either radian or degree measure. If we suppose that a scale
of inches and tenths is in the hands of the student and that a
graph is required upon an ordinary sheet of unruled paper of
letter size (8|- X 11 inches), then it will be convenient to let
1/5 inch of the horizontal scale of the X-axis correspond to 10°
or to ir/18 radians of angular measure. To accomplish this,
the length of one radian must be 1.15 inches (i.e., 18/5t inch),
which length must be used for the radius of the circle on which
the arcs of the angles are laid off. Hence, to graph y = sin x,
116 ELEMENTARY MATHEMATICAL ANALYSIS [§56
draw at the left of a sheet of (unruled) drawing paper a circle of
rddius 1.15 inches, as the circle OP5B, Fig. 59. Take 0 as the
origin and prolong the radius BO for the positive portion OX
of the X-axis. Divide OL into 1/5-uich intervals, each corre-
sponding to 10° of angle; eighteen of these correspond to the
length IT, if the radius BO (1.15 inches) be the unit of measure.
Next divide the F-axis proportionately to sin x in the following
manner : With a pair of bow dividers, or by means of a protractor,
divide the semicircle into eighteen equal divisions as shown
in the figure, thus making the length of each small arc exactly
1/5 inch. The perpendiculars, or ordinates, dropped upon OX
from each point of division, divided by the radius, are the sines
of the corresponding angles. Draw lines parallel to OX through
each point of division of this circle. ' These cut the F-axis at points
Ai, Ai, . Then if the radius of the circle be called unity,
the distances OAi, OA 2, OA3, . . are respectively the sines of
the angles OBPi, OBP2, OBP3, These are the successive
ordinates corresponding to the abscissas already laid off on OL.
The curve is then constructed as follows : First draw vertical lines
through the points of division of OX; these, with the horizontal
lines already drawn, divide the plane into a large number of rec-
tangles. Starting at 0 and sketching the diagonals (curved to
fit the alignment of the points) of successive "cornering" rec-
tangles, the curve OCNTL is approximated, which is the graph
oi y — sin x. This curve is called the sinusoid or sine curve.
The curve is of very great importance for it is found to be the
type form of the fundamental waves of science, such as sound
waves, vibrations of wires, rods, plates and bridge members,
tidal waves in the ocean, and ripples on a water surface. The
ordinary progressive waves of the sea are, however, not of this
shape. Using terms borrowed from the language of waves, we
may call C the crest, TV the node, and T the trough of the sinusoid.
It is obvious that as x increases beyond 27r'', the curve is re-
peated, and that the pattern OCNTL is repeated again and again
both to the left and to the right of the diagram as drawn. Thus
it is seen that the sine is a periodic function of period 2t' or 360°.
\For lack of room only a few of the successive points Pi, P2, P3, , , , of (iivigign
ef the quadrant OPjPf are actually lettered in Fig. 59,
§67] THE CIRCLE AND THE CIRCULAR FUNCTIONS 117
The small rectangles lying along the X-axis are nearly squares.
They would be exactly equilateral if the straight Hne OAi was equal
to the arc OPi. This equality is approached as near as we please
as the number of corresponding divisions of the circle and of OX
is indefinitely increased. In this way we arrive at the notion of
the slope of a curve in mathematics. In this case wfe say that the
slope of the sinusoid at 0 is + 1 and at A'^ is — 1, and at L is + 1.
We say that the curve outs the axis at an angle of 45° at 0 and
at an angle of 315° (or — 45° if we prefer) at N. The slope at C
and at T is zero.
The, curve y = a sin x is made from y = sin x by multiplying
all of the ordinates of the latter by a. The number a is called
the amplitude of the sinusoid.
57. The Cosine Curve. In Fig. 60 let the angles COPi, COP2,
COP3, etc., be laid off from the position of the F-axis OY as initial
side. Then if the radius of the circle be called unity, the dis-
Y
^Z),
0
D
D
¥-
s;
f^^ \ \
R/\ \ \ \
N
'
^
1
/\ \\\\
D.
S
/
/ ^\\\\
s
/
/ ^>A\\\\
s
/
\
A
L
/
A,
7r
\
or
/
H
TT
27r
?
S
/
?
\
\
/
\
s
/
/
_
_
_
_
—
—
—
^
—
—
—
^
—
_
_
—
—
T
Y
1
t-
Fig. 60. — Construction of the cosine curve.
tances ODj, OD2, OD3, . . are respectively the cosines of the
angles COP\, COP^, COP3, . If the distances laid off on the
Z-axis represent the measures of the successive angles COPi, COP2,
then the curve shown in the figure has the equation
y = cos X. The construction shows that the curve is exactly the
same as the sine curve of Fig. 59 except that the origin for the
cosine curve is under the crest while in the sine curve the origin
is at a node. If the origin be taken at 0' in Fig. 59 the curve may
be called the cosine curve.
118 ELEMENTAEY MATHEMATICAL ANALYSIS [§68
In Fig. 62 the curve ABODE is the cosine curve y = cos x. The
other curve is the sine curve y = sin x.
68. The Sine of a Negative Angle. In Fig. 61 the full drawn
curve represents the graph for y = sin x. The graph for
y = sin (— x) (1)
r
FiQ. 61. — The relation between y = aia x and y = sin (— s).
may be obtained by rotating the graph for y = sin x,
180° about the F-axis, by Theorem I on Loci. This gives the
dotted curve of Fig. 61. But from the properties of the sinusoid,
the dotted curve is the reflection in the Z-axis of the curve drawn
in full, hence the equation of the dotted curve may also be written
— 2/ = sin X. (2)
Hence, from (1) and (2)
sin (— x) = —sin x. (3)
69. Complementary Angles. Fig. 62 shows the curves for
y = cos X and for y = sin x. By the properties of these curves
Fig. 62. — Comparison of the sine and cosine curves.
it is obvious that the cosine curve may be regarded as the sine
curve translated x/2 units to the left. That is, the cosine curve
y = cos X (1)
has also the equation
y = sin (a; + I) • (2)
§60] THE CIRCLE AND THE CIRCULAR FUNCTIONS 119
Since this curve (the cosine curve) is symmetrical about the Y-
axis, its equation remains unchanged if we change x to {—x),
by Theorem I on Loci. Hence the cosine curve has also the
equation
y = sin ( - ^ + |\ = smi^-x\- (3)
Comparing (1) and (3) we see that we have proved for all values
of X that
sin ( x| = cosx. (4)
By comparing (1) with (2) we see that sin (s + *) ~ ^°^ ^'
this fact is, however, niuch less useful than that represented by
equation (4).
Exercises
From 'the curves for y = sin x and y = sin( — x), Fig. 61 shows
that:
1. sin {x — v) = sin {—x) and hence sin (tt — s) = sin x.
From the curves for y = cos x and y =^ sxixx, Fig. 62, shows that:
2. sin X = cos {x — s). 3. cos x = sin {x — fir).
4. cos {x + -fir) = sin x. 5. cos (.—x) = cos x.
60. Trigonometric Functions of Negative Angles. We have
already shown, (3) §57, that
sin (— x) = — sin x. (1)
Also from Fig. 62, since the cosine curve is symmetrical about the
y-axis,
cos (— x) = cos X. (2)
Dividing the members of (1) by the members of (2) we find
tan (— x) = tan x. (3)
61. Odd and Even Fimctions. A function that changes sign
but retains the same numerical value when the sign of the argu-
ment is changed is called an odd function. Thus sin x is an odd
function of x, since sin (—x) = —sin x. Likewise x^ is an odd
function of x, as are all odd powers of x. The graph of an odd
function of a; is symmetrical with respect to the origin 0 ; that is,
120 ELEMENTARY MATHEMATICAL ANALYSIS [§62
if P is a point on the curve, then if the line OP be produced
backward through 0 a distance equal to OP to a point P', then
P' also lies on the curve. The parts of 2/ = x' in the first
and third quadrants are good illustrations of this property.
A function of x that remains unaltered, both in sign and
numerical value, when the argument is changed in sign, is called
an even function of x. Examples are cos x, x', x^ — 3x*,
The graph of an even function is symmetrical with respect to the
y-axis.
Most functions are neither odd nor even, but mixed, like
x^ + sin X, x^ -\- x', and x + cos x.
Exercises
1. Is sin's an odd or an even function of x1 Is tan'a; an odd or
an even function of x7
2. Is the function sin x + 2 tan x an odd or an even'functfon? Is
sin X + cos x an odd or an even function of a;?
62. The Defining Equations Cleared of Fractions. The student
should commit to memory the equations defining the trigonometric
functions when cleared of fractions. In this form the equations
are quite as useful as the original ratios. They are written:
y = a sin 6 y = x tan 6 a = x sec 6
X = a cos 6 X = J cot d a = y esc 9
As applied to the right angled triangle, these three sets of equa-
tions may be stated in words as follows:
Either leg of a right triangle is equal to the hypotenuse multiplied
by the sine of the opposite, or by the cosine of the adjacent, angle.
Either leg of aright triangle is equal to the other leg multiplied by the
tangent of the opposite, or by the cotangent of the adjacent, angle.
The hypotenuse of a right triangle is equal to either leg multiplied
by the secant of the angle adjacent, or by the cosecant of the angle
opposite that leg.
These statements should be committed to memory.
63. Orthographic Projection. In elementary geometry we
learned that the projection of a given point P upon a given line or
plane is the foot of the perpendicular dropped from the given point
§63] THE CIRCLE AND THE CIRCULAR FUNCTIONS 121
upon the given line or plane. Likewise if perpendiculars be
dropped from the end points A and B of any line segment AB upon
a given line or plane, and if the feet of these perpendiculars be
called P and Q, respectively, then the line segment PQ is called the
projection of the line AB. Also, if perpendiculars be dropped
from all points of a given curve AB upon a given plane MO, the
locus formed by the feet of all perpendiculars so drawn is called the
projection of the given curve upon the plane MO.
To emphasize the fact that the projections were made by. using
perpendiculars to the given plane, it is customary to speak of them
as orthogonal or orthographic projections.
Pig. 63. — Orthographic projection of line segments.
The shadow of a hoop upon a plane surface is not the ortho-
graphic projection of the hoop unless the rays of light from the sun
strike perpendicular to the surface. This could only happen in
our latitude upon a suitable non-horizontal surface.
The shortening, by a given fractional amount, of a set of
parallel line segments of a plane may be brought about geometric-
ally by orthographic projection of all points of the line segments
upon a second plane. For, in Fig. 63, let AiBi, A^Bi, A3B3,
etc., be parallel line segments lying in the plane MN. Let their
projections on any other plane be Aid', AiC^', Ai'C-/, etc.,
respectively. Draw A\.C\ parallel to Ai'Ci' and Aid parallel to
122 ELEMENTARY MATHEMATICAL ANALYSIS [§63
Aj'Cj', etc. Then since the right triangles AiBiCi, AiB^Ct,
AiBaCz, etc. are similar,
AiBi A2B2 AsBs
AiCi A2C2 AiC
Call this ratio a. It is evident that a>l. Substitute the
equals: A/d' = AiCi, Aj'Ca' = A^d, etc. i'hen
AiBi _ A2B2 _ A3B3 _ , _a
Ai'Ci' ~ Ai'Ci' " As'c ~ ' ~r
The numerators are the original lii\e segments; the denominators
are their projections on the plane MO. The equality of these
fractions shows that the parallel lines have all been shortened in
the ratio a: I.
The above work shows that to produce the curve y = (x/a)",
(o < 1), from 2/ = a;" by orthographic projection it is merely neces-
sary to project all of the abscissas oi y = x" upon a plane passing
through YOY' making an angle with OX such that unity on OX
projects into a length a on the projection of OX. To produce the
curve y = ax" (a < 1) from y = x" hy orthographic projection it is
merely necessary to project all of the ordinates oi y = x" upon a
plane passing through XOX' making an angle with OY such that
unity on OY projects into the length a on the projection of OY.
To lengthen all ordinates of a given curve in a given ratio, 1 : a, the
process must be reversed; that is, erect perpendiculars to the plane
of the given curve at all points of the curve, and cut them by a^lane
passing through XOX' making an angle with OY such that a length
a (,a> 1) measured on the new K-axis projects into unity on OF of
the original plane.
In Fig. 50 the projection of OP in any of its positions, such as
OPi, OP2, OP3, ■ ■ ., is ODi, OD2, OD3, . . . , or is the abscissa of
the point P. Thus for all positions
X = a cos 6.
The sign of x gives the sign, or sense, of the projection. In each
case 0 is said to be the angle of projection.
This definition of projection is more general in one respect than
that discussed above. By the present definition the projection of
a line is negative if 90° < 9 < 270° (read, "if 6 is greater than 90°
§64] THE CIRCLE AND THE CIRCULAR FUNCTIONS 123
but is less than 270°"). This concept is important and essential
in expressing a component of a displacement, of a velocity, of an
acceleration, or of a force.
The cosine of 6 might have been defined as that proper fraction
by which it is necessary to multiply the length of a line in order to
produce its projection on a line making an angle d with it.
Exercises
1. A stretched guy rope 75 ft. long makes an angle of 60° with the
horizontal. What is the length of the projection of the rope on a
horizontal plane? What is the length of the projection of the rope
on a vertical plane?
2. Find the lengths of the projections of the line through the origin
and the point (1, -y/s) upon the OX and OF axes, if the Une is 12 inches
long.
3. A line 8 inches long makes an angle of 45° with the X-axis.
What is the length of its projection on the X-axis?
4. A velocity of 20 feet per second is represented as the diagonal
of a rectangle the longer side of which makes an angle of 30° with the
diagonal. Find the components of the velocity along each side of the
rectangle.
5. Show that the projections of a fixed hne OA upon all other
lines drawn through the point 0 are chords of a circle of diameter OA .
See Fig. 66.
6. Find the projection of the side of a regular hexagon upon the
three diagonals passing through one end of the given side, if the side
of the hexagon is 20 feet.
64. Polar Coordinates. In Fig. 64, the position of the point
P may be assigned either by giving the x and y of the rectangular
coordinate system, or by giving the vectorial angle 6 and the
distance OP measured along the terminal side of 6. Unlike
the distance o used in the preceding work, it is found conven-
ient to give the line OP a sense or direction as well as length;
such a line is called a vector. In the present Case, OP is known as
the radius vector of the point P, and it is usually symbolized by
the letter p. The vectorial angle 6 and the radius vector p are
together called the polar coordinates of the point P, and the
system used in locating the point is known as the system of polar
124 ELEMENTARY MATHEMATICAL ANALYSIS [§65
co5rdinates. In Fig. 64 the point P' is located by turning from
the fundamental direction OX, called the polar axis, through an
angle 6 and then stepping backward the distance p to the point
P'; this is, then, the point (— p, 9). P' has also the coordinates
(p, 02), in which 6^ = 0 + 180°; likewise Pi is (+ p', di) and
P'l is (— p', Bi). Thus each point may be located in the polar
system of coordinates in two ways, i.e., with either a positive or a
negative radius vector. If negative values of B be used, there
are four ways of locating a point without using values oi B>
360°. In giving a point in polar coordinates, it is usual to name
the radius vector first and then
the vectorial angle; thus (5, 40°)
means the point of radius vec-
tor 5 and vectorial angle 40°.
65. Polar Coordinate Paper.
Polar coordinate paper (form
MZ) is prepared for the con-
struction of loci in the polar
system. A reduced copy of a
sheet of such paper is shown
in Fig. 65. This plate is grad-
uated in degrees, but a scale of
radian measure is given in the
margin. The radii proceeding
from the pole 0 meet the circles at right angles, just as the two
systems of straight lines meet each other at right angles in rect-
angular coordinate paper. For this reason, both the rectangular
and the polar systems are called orthogonal systems of coordinates.
We have learned that the fundamental notion of a function implies
a table of corresponding values for two variables, one called the
argument and the other the function. The notion of a graph implies
any sort of a scheme for a pictorial representation of this table of
values. There are three common methods in use: the double scale,
the rectangular coordinate paper, and the polar paper. The polar
paper is very convenient in case the argument is an angle measured
in degrees or in radians. Since in a table of values for a functional
relation we need to consider both positive and negative values for
both the argument and the function, it is necessary to use on the
Polar coordinates.
§65] THE CIRCLE AND THE CIRCULAR; FUNCTIONS 125
polar paper the convention already explained. The argument, which
is the angle, is measured counter-clockwise if positive and clockwise
if negative from the line numbered 0°, Fig. 65. The function is
measured outward from the center along the terminal side of the angle
for positive functional values and outward from the center along the
terminal side of the angle produced backward through the center for
negative functional values. In this scheme it appears that four differ-
ent pairs of values are represented by the same point. This is made
FiQ. 65. — Polar coordinate paper.
clear by the points plotted in the figure. The points Pi, Pi, Pa, Pt
are as follows:
Pi: (6, 40°); (6, - 320°); (- 6, 220°); (- 6, - 140°).
Ps: (10, 135°); (10, - 225°); (- 10, 315°); (-10, - 45°).
Pa: (5, 230°); (5, - 130°); (- 5, 50°); (- 5, - 310°).
Pi-. {,&, 330°); (6, - 30°); (- 6, 150°); (- 6, - 210°).
The angular scale cannot be changed, but the functional scale
can be changed at pleasure.
In case the vectorial angle is given in radians, the point may
126 ELEMENTARY MATHEMATICAL ANALYSIS
be located on polar paper by means of a straight edge and the
marginal scale on form M3.
The point 0, Fig. 65, is called the pole and the line OA, the
polar axis.
Exercises
1. Plot upon polar coordinate paper the following: (a) (0.1, 30°;;
(6) (0.2,40°); (c) (0.6,120°); (d) (0.8,-30°); (e) (1.2,300°);
(/) (0.7, - 47°). Let 10 cm. = 1 unit for p.
2. Plot upon polar coordinate paper the following: (a) (1.3, 45°);
(6) (11.1, 137°); (cj (9.2, - 47°); (d) (8.5, - 216). Let 1 cm. = 1 unit
for p.
3. Plot upon polar coordinate paper the following: (a) (10, C);
(6) (9, D; (c) (8.2, 1.6'); (d) (12, 3.2"-). Let 1 cm. = 1 unit for p.
4. Explain why the locus for p = 3 is a circle with center at the
pole and radius equal to three units.
5. Draw the loci for p = 5 and p = 7.
6. Explain why the locus 8 = J tt is a straight line passing through
the pole and making an angle of 45° with the polar axis. Explain why
this locus is indefinite in extent and does not terminate at the pole.
7. Draw loci for: 9 = | tt, and d = — \ir.
8. Plot the locus for p = 9, if 9 is measured in radians. Use 2 cm.
as the unit for p.
66. Graphs of p = a cos 9 and p = a sin 0. These are two funda-
mental graphs in polar coordinates. The equation p = o cos 6
states that p is the projection of the fixed length a upon a radial
line proceeding from 0 and making a direction angle 6 with a,
or, in other words, p in all of its positions must be the side adjacent
to the direction angle 0 in a right triangle whose hypotenuse is the
given length o. (See §62 and Fig. 66.) It must be remem-
bered that the direction angle d is always measured from the fixed
direction OA. Hence, to construct the locus p = a cos 6, proceed
as follows: Draw a number of radical lines from 0, Fig. 66.
Project upon each of these the constant length OA, or a. These
projections are then radius vectors for p = a cos 0 and a curve
drawn through their end points gives the required locus.
Thi locus is a circle since P is always at the vertex of a right
triangle standing on the fixgd hypotenuse a, and therefore the
point P is on the semicircle AOP; for, from plane geometry, a right
triangle is always inscribable in a semicircle.
§66] THE CIRCLE AND THE CIRCULAR FUNCTIONS 127
When 6 is in the second quadrant, as 62, Fig. 66, the cosine is
negative and consequently p is negative. Therefore the point
P2 is located by measuring backward through 0. Since, however,
P2 is the projection of a through the angle 62 (see §63), the
angle at P2 must be a right angle. Thus the semicircle OP2A
is described as d sweeps the second quadrant. When 6 is in
the third quadrant, as ds, the cosine is still negative and p is
measured backward to describe the semicircle APiO a second
time. As 6 sweeps the fourth quadrant, the semicircle OP2A is
described the second time. Thus the graph in polar coordinates
Fig. 66. — The graph of p = a cos e.
of p = a cos d is a circle twice drawn as 6 varies from 0° to 360°.
Once around the circle corresponds to the portion ABC of the
"wave" y = a cos x, in Fig. 61. The second time around the
circle corresponds to the portion CDE from trough to crest of the
cosine curve. Trough and crest of all the successive "wave
lengths" correspond to the point A, the nodes to the point 0.
The polar representation of the cosine of a variable bj^ means
of the circle is more useful and important in science than the
Cartesian representation by means of the sinusoid. The ideas
here presented should be thoroughly mastered by the student.
The graph of p = a sin 6 is also a circle, but the diameter is
the line OB making an angle of 90° with OA, as shown in Fig. 67.
Since p = a sin 6, the radius vector must equal the side lying
opposite the angle 6 in a right triangle of hypotenuse a, if
128 ELEMENTARY MATHEMATICAL ANALYSIS ['§67
0° < e < 90°. Since angle AOPi = angle OBPi, the point Pi
is the vertex of any right triangle erected on OB, or a, as a hypote-
nuse. The semicircle BP2O is described as 0 increases from 90° to
180°. Beyond 180° the sine is negative, so that the radius vector
p must be laid off backward for such angles. Thus P3 is the
point corresponding to the angle 63 of the third quadrant. As 6
sweeps the third and fourth quadrants the circle OP1BP2O is
described a second time. Therefore the graph of p = asiad
is the circle tiince drawn of diameter a, and tangent to OA at 0.
The first time around the circle corresponds to the crest, the
second time around corresponds to the trough of the wave or
sinusoid drawn in rectangular coordi-
nates. 0 corresponds to the nodes of
the sinusoid and B to the maximum and
minimum points, or to the crests and
troughs.
We have seen that the graph of a
function in polar coordinates is a very
different curve from its graph in rect-
angular coordinates. Thus the cosine
of a variable if graphed in rectangular
coordinates is a sinusoid; but if graphed
FiQ. g7_ xhe graph of i^ polar coordinates it is a circle (twice
p = a sine. drawn) . There is in this case a very
great difference in the ease with which
these curves can be constructed; the sinusoid requires an elabo-
rate method, while the circle may be drawn at once with com-
passes. This is one reason why the periodic, or sinusoidal rela-
tion, is preferably represented in the natural sciences by polar
coordinates.
Exercises
1. Show that if — o is negative, p = — a cos 9 is a circle, diameter
a, with center to left of the pole §a units.
2. Show that if — a is negative p — — a sin 0 is a circle, diameter
o, with center below the pole 50 units.
67. Graphical Table of Sines and Cosines. The polar graphs
of p = a sin 0 and p = a cos 9 furnish the best means of construct-
ing graphical tables of sines and cosines. The two circles passing
§68] THE CIRCLE AND THE CIRCULAR FUNCTIONS 129
through 0 shown on the polar coordinate paper, form M3, Fig. 65,
are drawn for this purpose. A supply of this coordinate paper
should be in the hands of the student. If the diameter of the
sine and cosine circles be called 1, then the radius vector of any
point on the lower circle is the cosine of the vectorial angle, and
the radius vector of the corresponding point on the upper circle
, is the sine of the vectorial angle. Thus, from the diagram of
form M3, we read cos 45° = 0.707; cos 60° = 0.500; cos 30° =
0.866. These results are correct to the third place.
Exercises
1. /From coordinate paper, form M3, find the values of the following :
(a) cos 36°; (b) cos 62°; (c) cos 126°; (,d) sin 81°; (e) sin 25°; (/) sin 226°.
68. Graphical Table of Tangents and Secants. Referring to
Fig. 65, it is obvious that the numerical values of the tangents
of angles can be read off by use of the uniform scale bordering
the polar paper, form M3. The scale referred to lies just inside
of the scale of radian measure, and is numbered 0, 2, 4,
Thus to get the numerical value of tan 40° it is merely necessary to
call unity the side OA of the triangle of reference OAP, and then
read the side AP = 0.84; hence tan 40° = 0.84. To the same
scale (i.e., OA = 1) the distance OP = 1.31, but this is the secant
of the angle AOP, whence sec 40° = 1.31. By use of the circles
we find sin 40° = 0.64 and cos 40° = 0.76.
In case we are given an angle greater than 45° (but less than
135°) use the horizontal scale through B. Starting from B as
zero the distance measur/ed on the horizontal scale is the cotangent
of the given angle. The tangent is found by taking the reciprocal
of the cotangent.
Exercises
Find the unknown sides and angles in the following right triangles.
The numerical values of the trigonometric functions may be taken
from the polar paper. The vertices of the triangles are supposed
to be lettered A, B, C with C at the vertex of the right angle. The
small letters a, b, c represent the sides opposite the angles of the same
name. See also table of Natural Trigonometric Functions at end of
the book.
9
130 ELEMENTARY MATHEMATICAL ANALYSIS [§68
By angle of elevation of an object ia meant the angle between a
horizontal line and a line to the object, both drawn from the point of
observation, when the object lies above the horizontal line. The simi-
lar angle when the object lies below the observer is called the angle of
depression of the object.
The solution of each of the following problems must be cheeked.
The easiest check is to draw the triangles accurately to scale on form
Ml, measuring the unknown sides and angles.
1. When the altitude of the sun is 40°, the length of the shadow cast
by a flag pole on a horizontal plane is 90 feet. Find the height of the
pole.
Outline of Solution. Call height of pole a, and length of shadow
b. Then A = 40° and B = 50°. Hence,
o = 6 tan 40°.
Determining the numerical value of the tangent from the polar paper,
we find
a = 90 X 0.84 = 75.6 ft.,
which result, if checked, is the height of the pole. To check, either
draw a figure to scale, or compute the hypotenuse c, thus :
c = 90 sec 40°
From the polar paper find sec 40°. Then
c = 90 X 1.31 = 117.9
Since a^ + b' = c', we have c' - b^ = a'', or (c - 6) (c + 6) = a'.
Hence, if the result found be correct,
(117.9 - 90) (117.9 + 90) = (75.6) ^
5800 = 5715.
These results show that the work is correct to about three figures, for
the sides of the triangle are proportional to the square roots of the
numbers last given.
2. At a point 200 feet from, and on a level with, the base of a tower
the angle of elevation of the top of the tower is observed to be 60°.
What is the height of the tower?
3. A ladder 40 feet long stands against a building with the foot of
the ladder 15 feet from the base of the wall. How high does the ladder
reach on the wall?
4. From the top of a vertical cliff the angle of depression of a point
§68] THE CIRCLE AND THE CIRCULAR FUNCTIONS 13.1
on the shore 150 feet from the base of the cliff is observed to be 30°.
Find the heiglit of the cliff.
6. In walking halt a mile up a hill, a man rises 300 feet. Find the
angle at which the hill slopes.
If the hill does not slope uniformly the result is the average slope
of the hill.
6. A line 3.5 inches long makes an angle of 35° with OX. Find the
lengths of its projections upon both OX and OY.
7. A vertical cliff is 425 feet high. From the top of the cliff the
angle of depression of a boat at sea is 16°. How far is the boat from
the foot of the chff?
8. The projection of a line on OX is 7.5 inches, and its projection
on OY is 1.25 inches. Find the length of the line, and the angle it
makes with OX.
9.- A battery is placed on a cliff 510 feet high. The angle of depres-
sion of a floating target at sea is 9°. Find the range, or the horizontal
distance of the target from the battery.
10. From a point A the angle of elevation of the top of a monument
is 25°. From the point B, 110 feet farther away from the base of the
monument and at the same elevation as A, the angle of elevation is
15°. Find the height of the monument above the line AB.
11. Find the length of a side of a regular pentagon inscribed in a
circle whose radius is 12 feet.
12. Proceeding south on a north and south road, the direction of a
church tower, as seen from a milestone, is 41° west of south. From
the next milestone the tower is seen at an angle of 65° W. of S. Find
the shortest distance of the tower from the road.
13. A traveler's rule for determining the distance one can see from
a given height above a level surface (such as a plain or the sea) is as
follows : " To the height in feet add half the height and take the square
root. The result is the distance you can see in miles." Show that
this rule is approximately correct, assuming the earth a sphere of
raldius 3960 miles. Show that the drop in 1 mile is 8 inches, and that
the water in the middle of a lake 8 miles in width stands lOf feet
higher than the water at the shores.
14. Observations of the height of a mountain were taken at A and
B on the same horizontal line, and in the same vertical plane with the
top of the mountain. The elevation of the top at A is 52° and at B is
36°. The distance AB is 3500 feet. Find the height of the mountain.
16. The diagonals of a rhombus are 16 and 20 feet. Find the
lengths of the sides and the angles of the rhombus.
132 ELEMENTARY MATHEMATICAL ANALYSIS
16. The equation of a line is y = f.r +10. Compute the shortest
distance of this Une from the origin.
17. Find the perimeter and area of ABCD, Fig. 68.
18. Find BC and the total area of ABCD, Fig. 69.
69. The Law of the Circular Functions. It will be emphasised
in this book that the fundamental laws of exact science are three in
number, namely: (1) The power function expressed hy y = ax"
where n may be either positive or negative; (2) the harmonic or
periodic law y = aaia nx, which is fundamental to all periodically
occurring phenomena; and (3) a law to be discussed in a sub-
sequent chapter. While other important laws and functions
arise in the exact sciences, they are secondary to those expressed
by the three' fundamental relations.
Fig. 68. — Diagram for
Exercise 17.
Fig. 69. — Diagram for
Exercise 18.
We have stated the law of the power function in the following
words (see §34):
In any power function, if x change hy a fixed multipk, y
changes by a fixed multiple also. In other words, if x change by
a constant factor, y will change by a constant factor also.
Confining our attention to the fundamental functions, sine
and cosine, in terms of which the other circular functions can
be expressed, we may state their law as follows :i
1 Chapter XI is devoted to a diacuasion of theae fundamental periodic laws.
§70] THE CIRCLE AND THE CIRCULAK FUNCTIONS 133
The circular functions, sin 6 and cos B, change periodically in
value proportionally to the periodic change in the ordinate and
abscissa, respectively, of a point moving uniformly on the circle
a;2 + 2/2 = aK
The use of the periodic law in the natural sciences is, of course,
very different from that of the power function. The student will
find that circular functions similar toy = a sin nx will be required
in order to express properly all phenomena which are recurrent
or periodic in character, such as the motion of vibrating bodies,
all forms of wave motion, such as sound waves, light waves,
electric waves, alternating currents and waves on water surfaces,
etc. Almost every part of a machine, no matter how compli-
cated its motions, repeats its original motions at stated intervals
and these recurrent positions are expressible in terms of the
circular functions and not otherwise. The student will obtain
a very limited and unprofitable idea of the use of the circular
functions if he deems that their principal use is in numerical
work in solving triangles, etc. The importance of the circular
functions lies in the power they possess of expressing natural
laws of a periodic character.
70. Rotation of Any Locus. In §36 we have shown that any
locus y = f{x) is translated a distance a in the x direction by
substituting (x — a) for x in the equation of the locus. Likewise
the substitution of (y — b) for y was found to translate the locus
the distance b in the y direction. A discussion of the rotation of
a locus was not considered at that place, because a displacement
of this type is best brought about when the equations are ex-
pressed in polar coordinates.
If a table of values be prepared for each of the loci
p = cos 8 (1)
P = cos (9i - 30°) (2)
as follows:
134 ELEMENTARY MATHEMATICAL ANALYSIS [§70
Equation 1
Equation 2
e
p
9i
P
-30°
0.866
0°
0.866
-20°
0.940
10°
0.940
-10°
0.985
20°
0.985
0°
1.000
30°
1.000
10°
0.985
40°
0.985
20°
0.940
50°
0.940
30°
0.866
60°
0.866
40°
0.766
70°
0.766
60°
0.643
80°
0.643
60°
0.500
90°
0.500
and then if the graph of each be drawn, Fig. 70, it will be seen that
the curves differ only in location and not at all in shape or size.
If a value be given to 61
in the second equation which
is 30° greater than a value
given to d in the first equa-
tion, the two values of p
from equations (1) and (2)
are equal. Thus, if AOP is
the value given to S in equa-
tion (1) and if AOP' =
AOP + 30° is the value
given to Oi'va. equation (2),
then OP will equal OP'.
Thus the point P' may be
_,_„„. , , looked upon as having been
J)iG. 70. — Kotation of the circle „i . ; j f „ .i ■ i o
OAP [p =a cos 8] to the position obtamed from the pomt P
OA'P' \p = a cos (e — 30°)]. by a positive rotation about
0 of 30°. Thus the graph
for p = cos {d - 30°) may be obtained from the graph for p =
cos d by rotating it about the pole 0 through an angle of 30°.
The same reasoning will apply if {fi - a) be substituted for 6;
in this case the locus of the first curve is rotated about the pole
through an angle a, in the positive sense if a be positive, in a
negative sense if a be negative.
§71] THE CIRCLE AND THE CIRCULAR FUNCTIONS 135
By the same reasoning as used above, we see that if in the polar
equation of any curve, d is replaced by (6 — a), the graph of the
new equation is the graph of the original equation rotated about
the pole through an angle a, but is otherwise unchanged.
Thboeems on Loci
XIV. If {d — a) be substituted for 6 throughout the polar equa-
tion of any locus, the curve is rotated through the angle a.
Note that the rotation is positive when a is positive and nega-
tive when a. is negative.
Exercises
paper
draw: a = cos i
1. Upon a sheet of polar coordinate
p = cos (9 - 60°); p = cos (9 + 60°).
2. Upon a sheet of polar coordinate paper draw: p = sin B;
p = sin (.9 - 30°j; p = sin [6 '+ 30°).
3. Upon a sheet of polar coordinate paper draw: p = cos 9;
P = cos (fl — I) ; p = cos (9 + |j ; p = cos (9 — t).
71. Polar Equation of the Straight Line. In Fig. 71 let MN be
any straight line in the
plane and OT be the per- »^
pendicular dropped upon
MN from the pole 0. Let
the length of OT be a and
let the direction angle of
OT be a, where, for a
given straight line, a and
a are constants. Let p be
the radius vector of any
point P on the line MN
and let its direction angle
be d. Then, by definition,
- = cos (6 — or). Fig. 71. — ^Equation of MN is a = p cos
P (e-a).
Therefore the equation of the straight line MN is
a = p cos {d — a), (1)
136 ELEMENTARY MATHEMATICAL ANALYSIS [§72
for it is the equation satisfied by the (p, 6) of every point of the
line. This is the equation of any straight line, for its location is
perfectly general. The constants defining the line are the per-
pendicular distance a upon the given line from 0, and the direction
angle a of this perpendicular. The perpendicular OT, or a, is
called the normal to the line MN, and the equation (1) is called
the normal equation of the straight line.
The equation of the circle shown in the figure is
Pi = o cos {6 — a), (2)
in which pi represents the radius vector of a point Pi on the circle.
The relation pi p = a^, which can be deduced from (1) and (2),
is interesting. Because of it, the circle is often called the inverse
of the line MN with respect to the point 0.
Exercises
1. Write the polar equation of the line tangent tp the circle p =
5 cos (9 — 30°) at the end of the diameter passing through the pole.
2. A line is 3 units distant from the pole and makes an angle of
45° with the polar axis. Write its polar equation.
3. Describe the curves p = 10 cos I * — 4) and 10 = p cos ( ^ ~ i) •
Draw the following circles :
4.
p = 3 COE
1 (e - 30°).
7. P =
2 sin (6 + 135°).
6.
p = 3 cos
{e + 120°).
8. p =
^cos{e + l)-
6.
p = 2 sin
(9 - 45°).
9. p =
: 5 sin (1 - e) •
10.
Show that p
= a sin 9 is the locus p
= 0 cos 9 rotated 90° counter
clockwise.
72. Relation between Rectangular and Polar Coordinates.
Think of the point P, Fig. 72, whose rectangular coordinates are
(x, y). If the radius vector OP be called p and its direction angle
br I'alhd 0, then the polar coordinates of P are (p, 6). Then x and
y fo: any position of P are the projections of p through the
angle 6, and the angle (90° — d), respectively, or
X = p cos d, (1)
y = p sin 8. (2)
§72] THE CIRCLE AND THE CIRCULAR FUNCTIONS 137
These are the equations of transformation that enable us to write
the equation of a curve in polar coordinates when its equation
in rectangular coordinates is known, or vice versa. Thus the
straight line x = 3 has the equation
p cos 6 = 3
in polar coordinates. The line x + y = 3 has the polar equation
p cos d + p sin 6 = 3.
The circle x^ + y^ = a^ has the equation
p^cos" 9 + p2 sin^ e = o^
or
or
To solve equations (1) and (2)
for 8, we write
6 = the angle whose cosine is ->
P
6 = the angle whoser sine is -•
P
The verbal expressions "the
„„„!„ „,!,„„„ „„„; ;„ " „i„ Fig. 72. — Rectangular and polar
angle whose cosine is, etc., are coordinates of a point P.
abbreviated in mathematics by
the notations "arc cos," read "arc-cosine," and "arc sin," read
"arc-sine," as follows:
6 = aic cos (x/p) (3)
9 = arc sin (y/p) (4)
Dividing the members of (2) by the members of (1) we obtain
y
tan 0 = -J which, solved for 6, we write
0 = the angle whose tangent is
y
which may be abbreviated
d = arc tan (y/x)
and read "8 = the arc-tangent of y/x."
The value of p in terms of x and y is readily written
P = VxM^-
(5)
(6)
138 ELEMENTARY MATHEMATICAL ANALYSIS [§73
Exercises
1. Write the polar equation of x' + y' + 8x = 0.
The result is p* + 8p cos fl = 0, or p = — 8 cos e.
2. Write the polar equations of (a) x' + y^ — 4j/ =0; (6) a;' + y'
- 6x - iy = Q; (c) x^ +y'- Qy = 4.
3. Write the polar equations of {a) x + y = 1; (6) x + 2y = 1;
(c) X + Vly = 2.
4. Write the rectangular equations of (a) p cos 0 + p sin 9 = 4;
(6) p cos e — 3p sin 9 = 6.
5. Write the polar equation of x* + 2y' — 4x = 0.
6. Write the rectangular equation of p = 2 cos 9 + 3 sin 9.
Hint: Multiply both members of the equation by p, replace p' by
(x' + j/2), p cos 9 by X, and p sin 9 by y.
7. Write the rectangular equation of p = 3 cos 9 — 2 sin 9.
8. Write the rectangular equation of p = 5 sin 9 — 3 cos 9.
73. Identities and Conditional Equations. It is useful to make
a distinction between equalities like
(a - x){a + x) = a' - x\ (1)
which are true for all values of the variable x; and equalities like
x^ -2x = 3, (2)
which are true only for certain particular values of the unknown
number. When two expressions are equal for all values of the
variable for which the expressions are defined, the equality is
known as an identity. When two expressions are equal only for
certain particular values of the unknown number, the equality is
spoken of as a conditional equation. The fundamental formula
sin^ (j) + cos'' 0 = 1
is an identity.
2 sin A + 3 cos ^ = 3.55
is a conditional equation. The symbol = is sometimes used to
distinguish an identity; thus
a' — a;' = (a — x){a'^ -\- ax + x'').
The following illustrations and exercises contain problems both
in the establishment of trigonometric identities and in the finding of
the values of the unknown number from trigonometric conditional
equations.
§73] THE CIRCLE AND THE CIRCULAR FUNCTIONS 130
The truth of a trigonometric identity may be established by
reducing each side to the same expression. In this work, however, the
student will be required to transform the left-hand side by means of the
fundamental relations (2) to (6), §64, until it is identically equal to the
right-hand side.
Facility in the establishment of trigonometric identities is largely a
matter of skill in recognizing the fundamental forms and of ingenuity
in performing transformations. All solutions of conditional equations
should he checked. The following worked exercises will illustrate the
method.
Illustration 1: Show that (1 — sin u cos u) (sin u + cos u) ^
sin' u + cos' u. Taking the left-hand member
(1 — sin u cos u) (sin u + cos u)
= sin u + cos w — sin^ u cos u — sin u cos^ u
= cos u {1 — sin^ m) + sin u (1 — cos' u)
= cos u cos' u + sin u Bin' u
= cos' u + sin' u.
This last expression is the right-hand member of the given identity.
Thus the identity is verified.
Illustration 2: Show that sec' a; — 1 = sec' x sin' x.
sec' a; — 1 = see's (1 -; — ) = sec'x (1 — cos' x) = see's sin's.
\ sec' x/
Illustration 3: Solve for all values of x less than 360° '
2 sin X + cos s = 2.
Transposing and squaring we get
cos' X =4 — 8sina;-|-4 sin' x.
Since sin' x + cos' s = 1,
1 — sin' s = 4 — 8 sin x + 4 sin' x,
5 sin' a;-8sins + 3=0,
sin s = 1 or 0.6
X = 90°, and 37° or 143° approximately.
Check: 2 sin 90° -t- cos 90° = 2 -|- 0 = 2
Check: 2 sin 37° -f- cos 37° = 1.2 + 0.8 = 2
Does 2 sin 143° + cos 143° = 1.2 - 0.8 = 0.4 = 2?
The last value does not check. The reasons for this will be dis-
cussed later in §98. Therefore the correct solutions are 90° and 37°
approximately.
140 ELEMENTARY MATHEMATICAL ANALYSIS [§74
Exercises
1. Solve 6 cos" e + 5 sin 9 = 7 for all values of ff < 90°.
Suggestion: Write 6(1 — sin* 9) + 5 sin 9 = 7 and solve the
quadratic in sin 9.
6 sin" 9-5 sin 9 + 1 =0,
or
(3 sin 9 - 1)(2 sin 9 - 1) = 0.
sin 9 = i or J
0 = 19.6° approximately and 30°.
The results should be checked.
2. Prove that for all values of 9 (except ir/2 and 3ir/2, for which
the expressions are not defined)
sec* 9 — tan' 9 = tan* 9 + sec* 9.
3. Show that '
sec' u — sin* u = tan* u + cos* u,
for all values of the variable u except 90° and 270°, for which the
expressions are not defined.
4. Find u, if tan u + cot u = 2.
6. Find sec 9, if 2 cos 9 + sin 9 = 2.
6. Show that
sec a +1 tana
tan a ~ sec o — 1
Hint: Multiply both numerator and denominator of the left-hand
member by (sec a — 1).
7. Show that sec a + tan a = x
sec a — tan a
8. Show that sin* a + sin* a tan* a = tan* a.
9. Show that (esc* a — 1) sin* a = cos* a.
10. Show that
sin A _ 1 + cos A
1 — cos A ~ sin A
11. Show that 2 cos* m — 1 = cos* u — sin' u.
12. Show that cos' a — sin' a + 1 = 2 cos* a.
13. Show that sec* u + esc* u = esc* u sec* u.
14. Show that (tan a + cot o)* = sec* a esc* a.
16. Solve sec x — tan s + 1 = 0 for all values of x less than 360°.
74. The Graph of p = a cos 0 + b sin 0. Before reading this
section the student should review exercises 6 and 7, §72. Let us
§74] THE CIRCLE AND THE CIRCULAR FUNCTIONS 141
find the Cartesian equation for the curve whose polar equation is
p = a cos 0 + 6 sin 6, (1)
where a and b are any constants, positive or negative. First
multiply each member of (1) by p.
ap cos B + bp sin 6
(2)
Since p^ = x^ + y^, p cos B = x, and p sin 0 = y, equation (2)
beaomes
x^ + y^ = ax + by. (3)
Transposing and completing squares
[■-ir-b-i]'
+ b'
(4)
Fio. 73. — The circles p = a cos e, p = b sin $, and p = a cos 9 +
6 sin e, or the circles OA, OB, and OC respectively.
This is the Cartesian equation of a circle with center at the point
(io, ib) and of radius iy/a' + 6^. The circle passes through the
pole or origin since the coordinates (0, 0) satisfy the equation
(3), and also passes through the point (a, 6), since these coordiT
nates satisfy (3). Thus if upon the diameters of the circles p =
o cos 6 and|P = 6 sin B, we construct a rectangle, the circle having
a diagonal of this rectangle as a diameter, is the locus of p =
aoosd + b sin B. See Fig. 73.
142 ELEMENTARY MATHEMATICAL ANALYSIS [§75
Exercises
Draw the graphs for the following :
1. p = 2 cos 9 + 2 sin 9. 2. p = 3 cos 9 + 2 sin 9.
3. p = — 2 cos 9 + 2 sin 9. 4. p = — 3 cos 9 — 2 sin 9.
5. In Fig. 74 let a =2 and a = 30°. Find the equation of each of
the four circles in the form p = a cos 9 + 6 sin 9.
p=ac°^^
Fig. 74. — Diagram for Exercise 5.
75. Additive Properties. The shearing of a curve in a straight
line, considered in §38, may be thought of as the addition of the
ordinates of the curve and of the straight line, corresponding to a
given value of the abscissa. This sum is the corresponding
ordinate of the new curve. In the more general case the curve
y = fW) + P'(x) may be constructed from the curves y = f(x)
and y = F{x) by adding their ordinates. Thus the curve for
y = x^-\ — ) Fig. 75, was constructed by adding the ordinates of
the curves y = x^ and j/ = •
In the same way the curve for p = f(d) + F{6) may be con-
structed from the curves p = f{d), and p = F(,d) by adding the
radius vectors corresponding to the same value of the. vectorial
angle. Thus points on the circle p = 2 cos 6 + 3 sin 6, Fig. 73,
may be located by adding (using the bow dividers) the radius
§76] THE CIECLE AND THE CIRCULAR FUNCTIONS 143
vectors of p = 2 cos 9 and p = 3 sin 6. That is, OP = OPi + OPi
for all positions of OP.
I Exercises
1. Plot on polar coordinate paper the curve for p =3 cos 9 + 2 sin 9
making use of the circles p = 3 cos 9 and p = 2 sin 9.
2. Plot on polar coordinate paper the curve for p = cos 9+1
making use of the circles p = cos 9, and p = 1. Note that when
90° < 9 < 270°, the p for p = cos 9 is negative, and that the addition
referred to above is algebraic addition.
3. Plot upon polar coordinate paper the curve for p = 1 + sin 9,
making use of the circles p = 1, and p = sin 9.
4. Plot upon polar coordinate paper the curve for p = 2 cos 9 — 1.
,6. Plot upon polar coordinate paper the curve for p = cos 9 + 2.
76. Graph of y = tan x. If this graph is to be constructed on
a sheet of ordinary letter paper, 8^ inches X U inches, it is desirable
144 ELEMENTARY MATHEMATICAL ANALYSIS [§76
to proceed as follows:* Draw at the left of the sheet of paper a semi-
circle of radius 1.15 . . inches (that is, of radius = 18/5ir), so
that the length of the arc of an angle of 10°, or ir/18, radians will be
i of an inch. Take for the X-axis a radius COX prolonged, and
take for the F-axis the tangent OY drawn through 0, as in Fig.
r MA M' a'
/T\J lL_______
/A W \/
Q
rUj/T I _ _ _ E
k//^___l^\ /_\
T^ZA Z 5 _-_ ^Z ^^
/^^5Z ^ z S^ ^
l^^—/ \ / K ^
fe:cr— -P -^^•- ^'^ -il^ ^
^^^^ ' " "^ Z 3^ Z
wxSc ~^, Z \ ^ I
VV>^ _S^l "_Z___
^\ t :_j____
\\ / '\ / \
\ / '^ w
2T-"
Y' B N' B' N"
Fig. 76. — Graphical construction of the curve of tangents y =
tan X. For lack of room only a few of the points Si, St, ... Ti, Ti,
. . .are lettered in the diagram. The dotted curve is s/ = cot x.
76. Divide the semicircle into eighteen equal parts and draw radii
through the points of division and prolong them to meet OF in
points Ti, Ti, Ti, Tt, . . . Then on the F-axis there is laid off
a scale YY' in which the distances OTi, OT2, . . are propor-
' tional to the tangents of the angles OCSi, OCS2, . . . ; for the
tangents of these angles are OTi/CO, Orj/CO, . . . and CO is
the unit of measure made use of throughout this diagram. Draw
horizontal lines through the points of division on OF and vertical
> The student should understand the construction of Figs. 76 and 78, but it is
opt necessary that be actually draw them.
[§77] THE CIRCLE AND THE CIRCULAR FUNCTIONS 145
lines through J inch intervals on OX, thus dividing the plane
into a large number of small rectangles. Starting at 0, t, 2fir,
... — IT, — 2ir, . and sketching the diagonals of con-
secutive cornering rectangles, the curve oi y — tan x is approxi-
mated. Greater precision may be obtained by increasing as
desired the number of divisions of the' circle and the number of
corresponding vertical and hprizontal lines.
It is observed that the graph of the tangent is a series of similar
branches, which are discontinuous for x = ir/2, — ir/2, (3/2)ir,
— (3/2)ir, ... At these values of x the curve has vertical
asymptotes, as shown at AB, A'B', in Fig. 76.
If the number of corresponding vertical and horizontal lines
be increased sufficiently, the slope of the diagonal of any rectangle
gives a close approximation to the true slope of the curve at that
point.
It has already been noted that all of the trigonometric functions
are periodic functions of period 2v. It is seen in this case, how-
ever, that tan x has also the shorter period x; for the pattern
M'N' of Fig. 76 is repeated for^each interval ir of the variable x.
77. Graph of cot x. In order to lay off a sequence of values of
cot S on a scale, it is convenient to keep the denominator con-
P» Ps Pi
Pr, P. P^
Pi
^
^
7
m
\M
j£
4
^
^
Pii Z>io DsDtD, O DiDi Dz Di Di
Fig. 77. — Construction of a scale of cotangents.
stant in the ratio (abscissa) /(ordinate) which defines the cotangent.
The denominator may also, for convenience, be taken equal to
unity. Thus, in Fig. 77, the triangles of reference DiOPi, D2OP2,
for the various values of 6 shown, have been drawn so that
the ordinates P\Di, PzDi, . are equal. If the constant ordi-
nate be also the unit of measure, then the sequence ODi, OD2, OD3 ,
ODt, ODi, represents, ' in magnitude and sign, the cotan-
gents of the various values of the argument d. Using ODi, OD2,
... as the successive ordinates and the circular measure of Q
10
146 ELEMENTARY MATHEMATICAL ANALYSIS [§78
as the successive abscissas, the graph oi y — cot x is drawn, as
shown by the dotted curve in Fig. 76.
The sequence ODi, OD2, Fig. 77 is exactly the same as
the sequence OTi, OT2, . Fig. 76, but arranged in the
reverse order. Hence, the graph of the cotangent and of the
tangent are alike in general form, but one curve descends as the
other ascends, so that the position, in the plane ZF, of the branches
of the curve are quite different. In fact, if the curve of the
tangents be rotated about 07 as axis and then translated to the
right the distance ir/2, the curves would become identical.
Therefore, for all values of x,
tan (7r/2 — x) = cot x. (1)
This is a result previously known.
78. Graph of y = sec x. Since sec 6 is the ratio of the radius
divided by the abscissa of any point on the terminal side of the
angle d, it is desirable, in laying off a scale of a sequence of values
of sec d, to draw a series of triangles of reference with the abscissas
in all cases the same, as shown in Fig. 78. In this figure the angles
were laid off from CQ as initial line. Thus
CTs/CSi = sec QCSi,
or, if CSi be unity, the distances like CTt, laid off on CQ, are the
secants of the angles laid off on the arc QSi,0 or laid off on the axis
OX.
The student may describe the manner in which the rectangles
made by drawing horizontal lines through the points of division on
CQ and the vertical Unes drawn at equal intervals aloiig OX, may
be used to construct the curve. If the radius of the circle be 1.15
inches, what should be the length of Oir in inches?
The student may sketch the locus oi y = esc x, and compare
with the locus y — sec x.
Exercises
1. Discuss from the diagrams, 59, 76, 78, the following statements:
Any number, however large or small, is the tangent of some angle.
The sine or cosine of any angle cannot exceed 1 in numerical value.
The secant or cosecant of any angle is always numerically greater^
than I (,or at least equal to 1),
§79] THE CIRCLE AND THE CIRCULAR FUNCTIONS 147
2. Show that sec (o ~ ^) ~ <'*° ^ ^°'' ^^ values of x.
3. If tan 9 sec 9 = 1, show that sin $ = KVs — 1) and find 9
by use of polar coordinate paper, Form M3.
4. Describe fully the following, locating nodes, troughs, crests, etc. :
(a) y = sin [x -"^y (c) y = tan y> +lj'
(b) y = cos (^J + 1) ' (d) y = tan (x + 1).
r
N
A
N'
\\
/
1
\
T
\\\
/
^
\ \ \ \
/p
S '~
WW \
/
_
._
^
«
®„
m
7
r
M
2'
'///I
7
<
"
///
/
\
/■/ /
/
\
/ /
/
\
I" B N' , N'
Fig. 78. — Graphical construction of y' = sec x.
79. Increasing and Decreasing Functions. The meanings of
these terms' have been explained in §27. Applying these terms to
the circular functions, we may say that y = sin x, y = tan x,
y = sec X are increasing functions for 0 < a: < ■k/2. The co-
functions, y = cos z, y — cot x, y = esc x, are decreasing func-
tions within the same interval.
148 ELEMENTARY MATHEMATICAL ANALYSIS [§79
Exercises
Discuss tlie following topics from a consideration of the graphs of
the functions:
1. In which quadrants is the sine an increasing function of the
angle? In which a decreasing function?
2. In which quadrants is the tangent an increasing, and in which a
decreasing, function of its variable?
3. In which quadrants are the cos 0, cot 6, sec 9, esc 6, increasing
and in which are they decreasing functions of 9?
4. Show that all the co-functions of angles of the first quadrant are
decreasing functions.
1. Show that
2. Show that
Miscellaneous Exercises
tan' a . ,
T — r^ — r- — sin' a
1 + tan' a
\/l — sin'a COS a
y/\ — COS
3. Show that cot' a — cos' a = cot' a cos* o.
4. Show that
s Vcsc' - 1*
6. Show that
6. Show that
7. Show that
Vsec'a — 1
1 + tan' a __ sin' a
1 + cot' a. " cos' a
1 + cos a
8. Show that
CSC a
\\ — sin a ,
\'- — - — , = sec a — tan a.
I + sin a
sin a , 1 + cos a
-\ ; = 2 CSC a
COS a.
cot a + tan a
9. Show that
1
— T r-L = sin u cos M.
cot u + tan u
10. Show that
CBO* M (1 -r cos* m) — 2 cot' Mai.
§79] THE CIRCLE AND THE CIRCULAR FUNCTIONS 149
11. Find the distance of the end of the diameter of ,
p = 8 cos (9 - 60°)
from the line OX.
12. If PI = a cos 9, and P2 = a sin 6, find pi — pi when 0 = 60°
and 0=5.
13. Find the polar equation of the circle x' + y' + Qx = 0:
14. For what value of 9 does p = 3.55, if p = 2 sin 9 + 3 cos 9?
Result: 9 = 23° 30' and 43° 30'. Hint: Draw the circles p = 3.55
and p = 2 sin 9 + 3 cos 9 on polar coordinate paper and find the
vectorial angles for the two points of intersection. This problem is
the same as: "Solve the equation 2 sin 9 + 3 cos 9 = 3.55 for 9."
16. Solve graphically the equation 2 sin 9 + 3 cos 9 = 2.
Hint: Draw on polar coordinate paper the curves p = 2 and
p = 2 sin 9 + 3 cos 9.
16. Solve graphically the equation 4 cos 9 — 3 sin 9= 3.5.
17. Find sin 9 if esc 9= vV_±A'.
a
18. A circular arc is 4,81 inches long. The radius is 12 inches.
What angle is subtended by the arc at the center? Give result in
radians and in degrees.
19. Certain lake shore lots are bounded by north and south lines
66 feet apart. How many feet of lake shore to each lot if the shore-
line is straight and runs 77° 30' E. of N.7
20. If 2/ = 2 sin A + 3 cos A - 3.55, take A as 20°; as 23°; as 26°
and find in each case the value of y. From the values of y just
found find a value of A for which y is approximately zero. This
process is known as "cut and try."
21. The line y = ^x ia to coincide with the diameter of the circle
p = 10 cos (9 — a). Find a.
22. The line y = 2x is to coincide with the diameter of the circle
p = 10 sin (9 + a). Find a.
23. To measure the width of the slide dovetail shown in Fig. 79,
two carefully ground cylindrical gauges of standard dimensions are
placed in the V'a at A and B, as shown, and the distance X carefully
taken with a micrometer. The angle of the dovetail is 60°. Find
the reading of the micrometer when the piece is planed to the required
dimension MN = 4 inches. Also find the distance Y. (Adapted
from "Machinery," N. Y.)
24. Sketch y = ix and y = sinx and then y = ix — sinx.
26. Sketch the curve y = cos .-c + 2 sin x, making use of the curves
y = cos X and ^ = 2 sin x.
150 ELEMENTARY MATHEMATICAL ANALYSIS [|79
26. Find the maximum value of the function given in exercise 25.
Hint: Find the maximum value of p in the graph of p = cos 9
+ 2 sin e.
27. Find the maximum value of 2 cos x — 3 sin x.
28. Since p = cos 9 + 2 sin fl is a circle passing through the pole,
the equation may be put in the form p =' a cos (9 — a). Find a
and a.
Result: o = VS and a = 63° 20' approximately.
29. A circle is inscribed in a 30°, 60° right triangle. Find the diame-
ter of the circle (a) if the shorter leg of the triangle is 2 inches; (6)
if the longer leg is 2 inches; (c) if the hypotenuse is 4 inches, (d) Find
the length of the sides of the triangle if the diameter of the inscribed
circle is 2 inches.
Fig. 79. — Diagram to Exercise 23.
30. A circle is inscribed in a 45° right triangle. Find the diameter
of the circle if the legs of the triangle are 4 inches.
31. The center of the circle p'= 10 cos (9 — a) Ues on the Ijne
Zx — 2y = \. Find two possible values for a.
32. The center of the circle p = 10 sin (9 + a) lies on the line
X — 22/ = 6. JFind two possible values for a.
33. Write the Cartesian equations for:
(a) p = 2 cos 9 + 3 sin 9. (b) p = 2 cos 9 — 5 sin 9.
(c) p = 2 sin 9 — 5 cos 9.
34. Find the co6rdinates of the center and the radius for:
(o) x» + 2/2 - 2x - 4i/ + 4 = 0 (d) 2x2 + 2y^ + 3x + by= 0
(5) x« + 2/' + 2x + 42/ + 4 = 0 (e) 3x2 + ^yi _ gj. - ■y/2y = 10
(c) x' + !/» + 3x - 42/ = 0 (/) x2 + ^2 + 7x- \^Zy = 25
§79] THE CIRCLE AND THE CIRCULAR FUNCTIONS 151
36. Which circles of exercise 34 pass through the origin?
36. Write the equation of a line passing through the origin and the
center of the circle x' + y^ — 3x — 5y = 6.
37. Write the equation of a Une parallel to3x —2y = Q and passing
through the center of x' + y'' — Sx — 2y = 0.
CHAPTER V
THE ELLIPSE AND HYPERBOLA
80. The Ellipse. If all ordinates of a circle be shortened by
the same fractional amount of their length, the resulting curve
is called an ellipse. For example, in Fig. 80, the middle points
of the positive and negative ordinates of the large circle were
marked and a curve drawn
through the points so selected.
The result is the ellipse
ABA'B'A.
If
a;2 + 2/2 = a* (1)
is the equation of a circle, then
x^ + {myY = o^ (2)
in which m is any constant > 1,
is the equation of an ellipse;
for substituting my for y divides
all of the ordinates by m, by
Theorem IX on Loci, §28.
Fig. 80.-
-Construction of an
ellipse.
Dividing both members of (2) by a' we obtain.
-i + '-iV
= 1.
(3)
Let 6' be written in place of — ^- Equation (3) becomes
m'-
V
+ h = ^
(4)
which is the standard form of the equation of an ellipse.
81. Orthographic Projection of a Circle. The ellipse may
also be looked upon as the orthographic projection of the circle.
152
§81]
THE ELLIPSE AND HYPERBOLA
153
Let ABCD, Fig. 81a, be a circle with a radius o. Let AOC, Fig.
816, be an end elevation of the same circle. Rotate this circle
about BOD as an axis through an angle, /J, to the position A"OC".
Project the rotated circle upon its original plane, into the curve
A'BC'D. We shaU show that A'BC'D is an ellipse. Take any
point P upon the original circle. It rotates into the point P", and
P" projects into P'. The equation of the circle is x^ + y^ = o",
where y = MP. To get the equation for the curve A'BC'D
replace MP by its equal MP'/cos /?. (See Fig. 81b.) Whence,
' +
(MP'y
cos''/3
Since MP' is the ^/-coordinate of P',
x^ +
y2
cos'/3
A
P /
]
^
p
P' /^
~~?'
^
\
n ^\
o
V
a
^\ VV
/
/
y
b " ~ a
Fig. 81. — The ellipse considered as the orthographic projection of
a circle.
or
V'
o' a^oos'jS
Replacing o cos jS by 6(= OC),
the equation of an ellipse.
1,
As a consequence of the above, it ia seen that the shadow cast on
164 ELEMENTAKY MATHEMATICAL ANALYSIS [§82
the floor by a circular hoop held at any angle in the path of vertical
rays of light is an ellipse.
If the abscissas of a circle be lengthened by amounts propor-
tional to their lengths, the resulting curve is an ellipse. Let
x^ + y^ — V be the equation of the circle. Then
2
+ 2/2 = b2
is the equation formed by lengthening all abscissas in the raticf
1 : m, TO > 1. Dividing by Ji^ and replacing m'-h'^ by a}, we obtain
a^ "^ 62 ^•
Thus the ellipse of Fig. 80 could have been formed by doubling all
of the abscissas of the circle BD'B'D. Hence we see that if all
parallel chords of a circle are lengthened or shortened by an
amount proportional to their length, an ellipse is formed. If the
deformation takes place in chords parallel to either the X- or
F-axis the equation is of the form (4) of the last section, called the
symmetric equation of the ellipse.
The diameter of the circle from which the ellipse may be formed
by shortening parallel chords is called the major axis of the
ellipse. Thus AA', or 2a, Fig. 80 is the major axis of ABA'B'.
The diameter of the circle from which the ellipse could have been
formed by lengthening parallel chords is called the minor axis of
the ellipse. Thus BB', or 26, Fig. 80 is the minor axis of ABA'B'.
The point of intersection of the axes is called the center of
the ellipse. One-half of the major and minor axes are called,
respectively, the semi-major and semi-minor axes of the ellipse.
The points A and A' are called the vertices of the ellipse.
82. Ejcplicit Form of Equation. The equation of the ellipse
o" ^ 62 ^'■'
when solved for y may be put in the important form
.y = + - Va^-x^ (2)
The equation of the circle x^ + y^ = a^ solved for y is
y = ± Va^ - x2- (3)
§83]
THE ELLIPSE AND HYPERBOLA
155
Equations (2) and (3) are in a form very useful for many purposes.
It is easy to see that (2) states that the ordinates of the ellipse are
the fractional amount hfa of the ordinates of the circle -(3) .
The definition of the term function permits us to speak of j/ as a
function of x, or of a; as a function of y, in cases like equation (1)
above; for when x is given, y is determined. To distinguish this
from the case in which the equation is solved for y, as in (2), y, in
the former case, is said to be an implicit function of x, and in the
latter case, y is said to be an explicit function of x.
83. Section of a Cylinder. If a circular cylinder be cut by a
plane, the section of the cylinder is an ellipse. For, select any
diameter of a circular section of
the cylinder as the X-axis. Let
a plane be passed through this
diameter making an angle a
with the circular section. Then
if ordinates (or chords perpen-
dicular to the common X-axis)
be drawn in each of the two
planes, all ordinates of the sec-
tion made by the cutting plane
can be made from the ordinates
of the circular section by multi-
plying them by sec a. Hence
any plane section of a cylinder
is an ellipse.
84. Parametric Equations of the Ellipse. Let ABA'B'A, Fig.
82, be an ellipse whose semi-major axis is a and whose semi-minor
axis is 6. Upon AA' and BB' as diameters construct circles.
These circles are called, respectively, the major and minor auxiliary
circles. From the origin, draw any radius vector, as QP2P1,
making an angle d with the positive direction of the axis of x.
Through P2 and Pi draw lines parallel, respectively, to the X- and
F-axes, and let P be their point of intersection. It will be shown
that P is a point upon the ellipse.
Let the coordinates of P be a; and y. Then
-A construction of the
ellipse.
X = a cos
(1)
156 ELEMENTARY MATHEMATICAL ANALYSIS [§84
and
or
and
Then
y = b sin 6, (2)
- = cos 8,
a
f =sine.
0
^ + f" = cos^ fl +sin2 e = 1,
which shows that P is upon the ellipse.
Equations (1) and (2) are called parametric equations of the
ellipse. 6 is called the variable parameter, or the eccentric angle.
The method used above of locating points upon the ellipse
constitutes one of the best practical methods of constructing an
ellipse when its axes are known. For, by it a large number of
points upon the ellipse may be easily, located and a smooth curve
drawn through them.
If the abscissa and ordinate of any point of a curve are ex-
pressed in terms of a third variable, the pair of equations are
called the parametric equations of the curve. Thus,
X = U
y = t + l
are the parametric equations of a certain straight line. Its
ordinary equation
y = ix + l
can be found by eliminating the parameter t.
Exercises
1. Write the equation of the ellipse formed by diminishing the
lengths of all ordinates oix' + y^ =4 by one-half of their length.
2. Write the equation of the ellipse formed by diminishing the'
lengths of all ordinates oi x' + y' =4 by one-third of their lengths.
3. Write the equation of the ellipse formed by lengthening all
ordinates of the circle x' + y^ = 16 by one-third of their length.
4. Write the equation of the ellipse formed by lengthening all
abscissas of the circle x' + ^' = 1 by one-fourth of their length.
§84] THE ELLIPSE AND HYPERBOLA 157
6. Write the equation of the ellipse whose semi-axes are 4 and 3.
6. Construct accurately an ellipse whose semi-axes are 3 inches and
2 inches.
7. Construct accurately an ellipse whose parametric equations are
x = 3 cos e, and y = 2sm e.
8. Write the parametric equations of an ellipse whose semi-axes
are 6 and 10.
9. Draw a curve whose parametric equations are x = cos 6, and
y = sin e.
10. Find the major and minor axes for the following :
(c) 4x' + 25y' = 100 (d) 25a;2 + 4y' = 100
11. Find the axes of the ellipse k'x^ + h'y^ = hV.
12. Write the equation of the ellipse whose major and minor axes
are 10 and 6, respectively.
13. Find the axes of the elUpse whose equation is
2/ = ± i V36 — a;^ [Note that o must be 6.]
14. Write the parametric equations of the ellipse
y = + IVSl -x2. [a must be 9; b = f X 9 = 6.]
16. Discuss the curve
a; = ± I v'4 - 2/2.
16. Discuss the following curves by comparing them with
a;2 + 2/" = 1-
4x2 -1-2/2 = 1
\x^ H- 2/2 = 1.
17. Write the Cartesian equation of the curves whose parametric
equations are:
. , Fa: = 2 cos 9 , . Tx = 6 cos 6 i \ \^ ~ V^^ cos 9
^°'' ly = sin e ^ ' ly =2 sin S ^'^' ly = V2 sin B.
18. What locus is represented by the parametric equations
X = 2t + 1
2/ = 3« -f- 5?
19. What curve is represented by the parametric equations
X = 2 -H 6 cos e
and 2/ = 5 -1- 2 sin e?
158 ELEMENTARY MATHEMATICAL ANALYSIS
20. Show that the curve
X = 3 + 3 cos e
2/ = 2 + 2 sin 9
is tangent to the co6rdinate axes.
Rg. 83.— a mechanical cons. ru.tion of the ellipse. See Exercise 23.
21. The circle x' + y^ = 36 is picjocted upon a plane. Find an
equation of the projection if the angle between the plane and the
plane of the circle is 30°.
22. A right circular cylinder is cut by a plane making an angle of
60° with the axis of the cylinder. Find an equation of the curve of
intersection, if the radius of the cylinder is 6 units.
Fig. 84. — Theory of the common "ellipsograph" or elUptic trammel.
See Exercise 24.
23. The line AB, Fig. 83, whose length is (a + 6) moves in such a
way that the ends A and B always lie on the X- and K-axes, respect-
ively. Show that the point P describes an ellipse.
24. The- edge of a straight ruler, NMP, Fig. 84, is marked so that
§84]
THE ELLIPSE AND HYPERBOLA
159
PM = b and PN = a. It is moved keeping M and N always on
AA' and BB', respectively. Show that P describes an ellipse. The
elliptic "trammel" or "ellipsograph" is constructed on this principle
by use of adjustable pins on PMN and grooves on AA' and BB'.
25. Draw a semicircle of radius a about the center C, Fig. 85, and
produce a radius to 0 such that CTO = a + 6. From C draw any
number of lines to the tangent to the circle at T. From 0 draw hnes
meeting the tangent at the same points of TN. At the points where
the lines from C cut the semicircle, draw parallels to CT. Show that
Fig. 85. — A graphical construction of an eUipse. See Exercise 25.
the points of meeting of the latter with the lines radiating from O
determine points on an ellipse, with center at 0 and semi-axes equal
to a and b.
Hint ; OD = SM = a cos 9. From the triangles OPD and ONT
OD _PD
TN b
From the triangles CNT and CMS
SM{=OD) _ SC{==a sine)
TN ~ a '
160 ELEMENTARY MATHEMATICAL ANALYSIS [§85
Hence
PD =b sin e.
86. Origin at a Vertex. Equation (4) §80, equation (2) §82, and
equations (1) and (2) §84 are the most useful forms of the equa-
tions of the ellipse. It is obvious that the ellipse may be
translated by the methods already explained to any position in
the plane. The ellipse with center at (A, k) and its axes parallel
to the coordinate axes has the equation
{x - hy {y - kr . ,.s
a2 -r 52 - ^' ^^)
Of special importance is the equation of the ellipse when the origin
is taken at the left-hand vertex. This form is best obtained from
equation (2), §82, by translating the curve the distance a in the
X direction. Thus,
y=± -Vo» - (x - ay,
u
or
y^ = — X -„ x^,
a a^ '
or, letting 21 stand for the coeflBcient of x,
2/2 = 2lx -~^x^ = 2lxil - x/2a). . (2)
For small values of x, x/2a is very small as compared with 1 and
the ellipse nearly coincides with the parabola y^ = 2lx.
86. Theorem. Any equation of the second degree, lacking the
term xy and having the terms containing x^ and y' both present and
wUh coefficients 0/ like signs, represents an ellipse with axes
parallel to the coordinate axes. This is readily shown by putting
the equation
ax^ + hy' + 2gx + 2fy + c = 0 (1)
in the form (1) of the preceding section. The procedure is as
follows:
a{x' + 2^^x) +h[y' + 2f^y)^-c. (2)
a(x^ + 2lx + Q +&(2/^ + 2(. + |) = f + f-c. (3)
§87] THE ELLIPSE AND HYPERBOLA 161
Let M stand for the expression in the right-hand member of (3) ;
then we get
a b
This shows that (1) is an ellipse whose center is at the point
and which is constructed from the circles whose cen-
\ a' hi
ters are at the same point and whose radii are the square roots of
the denominators in (4). The major axis is parallel to OX or
OY according as a is less or greater than 6. The case when the
locus is not real should be noted. Compare §43.
iLLtrsTRATiON: Find the center and axes of the ellipse
a;2 _|- ^yi _|_ 6j; _ 8?/ = 23.
Write the equation in the form
x'^ + Qx + 4j/2 -Sy = 23.
Complete the squares
a" + 6x + 9 + 4y2 - 8?/ + 4 = 36.
Rewriting (a + 3)' + 4(!/ - l)' = 36.
(x + 3)' , {y - ly ^
36 "^ 9
This is seen to be an ellipse whose center is at the point (—3, 1) and
whose semi-axes are 6 and 3.
87. Limiting Lines of an Ellipse. It is obvious from the
equation
y = +-^/o''-x^
that a; = a and x =. — a are limiting lines beyond which the curve
cannot extend; that is, x cannot exceed o in numerical value
without y becoming imaginary. The same test may be applied
to equations of the form
a;2 + 4x -h 92/2 - 6?/ -(- 4 = 0.
162 ELEMENTARY MATHEMATICAL ANALYSIS [§87
Solving for y in terms of x
3y = l + Vl-{x + 2)2.
The values of y become imaginary when
(x + 2)2>I,
or
a: + 2>+lor<-l,
or
x> - 1 or < - 3.
These, then, are the limiting lines in the x direction. Finding
the limiting lines in the y direction in the same way, the rectangle
within which the ellipse must lie is determined.
In cases . like the above the actual process of finding the limiUng
lines and the location of the center of the ellipse is best carried out
by the method of §86.
Illustration: Find the coordinates of the center, the length of the
axes, and the equation of the limiting lines of
x' + ix + Qy^ - Qy = 4.
Completing the squares,
(X + 2y + 9(2/ - \Y = 9,
or
(X + 2)' (y - \y _
g— + J 1-
The center of the ellipse is at the point ( — 2, |), its semi-axes are 3 and
1. It may be constructed by translating the ellipse -„ + y = 1, two
units to the left and \ unit up. Hence the limiting lines are
a; = + 3 — 2 and v = + 1 + 3, or x = 1, x = — 5, ?/ = f , and
y--l
Exercises
Find the lengths of the semi-axes, the coordinates of the center, and
the equations of the limiting lines for the seven following loci and
translate the curves so that the terms in x and y disappear, by the
method of §87.
1. x^ - 6x -I- 42/2 + 82/ = 5.
2. 2/2 - 8i/ + 4x2 -h 6 = 0.
§87] THE ELLIPSE AND HYPERBOLA 163
8. 12x« - 48a; + 3y' + 6y - 13.
4. x^ + %2 - 12a; + 6y =■ 12.
5. 4a;« + y' -12x + 12 2/ - 2 = 0.
6. x' +2y' - X - V2y = 1/2.
7. Show that a;^ — 4a; + 4^^ ^ 82/ + 4 = 0 is an ellipse.
8. Show that x' + 4a; + Qy' — 6y = 0 passes through the origin.
9. Discuss the curves :
,.,£■+!! = , wM:-'
10. Compare the following parabolas with the standard parabola
y = a;2 by means of the appropriate Theorems on Loci:
k
(a) y = 2px2 (c) j/ = — x^
(h) y 2pa;2 (d) y 2px^ + 6.
What are the roots of the last function?
11. Write the symmetrical equation of the ellipse if its parametric
equations are:
X = (3/2) cos e
y = (2/3) sin e.
12. Discuss the curve y^ = (18/5)x - (9/25)x2.
13. Find the center of the curve y^ = 2x (6 — x).
14 Write the parametric equations for the following :_
(a) x^ + 3y' =4; (6) 2x^ + 5y^ = 6; (c) 5x= + y^ = 7.
15. Write the parametric equations for
x' +2x + 4;/2 - l&y + 13 = 0.
Hint ; The equation may be put in the form
(x + ir , (y - 2)'
Since x = 2 cos 6 and y = am. 9 are parametric equations for
x^ v'^
-7 +Y = 1, x=2cosfl — 1 and 2/ = sin 9 + 2 are parametric
equations of the given ellipse.
16. Write parametric equations for the following : (a) x' — 2x +
9yi - 6x = 0; (6) 4x2 + 4x + j/2 - 2?/ = 5. (c) 3;2 _ 4^ ^ ^2 + gj/ = 3.
164 ELEMENTARY MATHEMATICAL ANALYSIS
88. The Rectangular Hyperbola. In §34 the graph of xy =■ k,
where fc is a constant, was called a rectangular, or equilateral,
hyperbola. It was observed that the X- and F-axes are asymp-
totes of the curve. We shall now find the equation of the equi-
lateral hyperbola when rotated about the origin through an
angle of ( — 45°). For convenience let k be represented by jo^.
Since this is a positive number, the curve will appear in the first
and third quadrants, as shown by the curve RPS, Fig. 86.
Fig. 86. — The rectangular hyperbolas 2 xy — a' and x' — y' = a'.
Let P be any point on the original curve
2xy = aK (1)
Let P' be this poiat after rotation. Let OD' = x and let B'P' = y.
OB = EP = E'P' = D'K - D'H = OD' cos 45° - D'P cos 45°
= iV2 {X - y). (2)
DP = OE = O'E' = OK + KE' = OD' cos 45° + D'P cos 45°
^W2(.x + y). , (3)
But OD is x and DP is y in equation (1). Substituting then
§89] THE ELLIPSE AND HYPERBOLA 165
l\/2(,x — y) and i\/2{x + y) for x and y, respectively, in equa-
tion (1), we obtain
x2 - y2 = a2, (4)
the equation of the curve R'P'S', or the equation of the rectangu-
lar hyperbola 2xy = a^ after it has been rotated (— 45°) about
the origin.
The equation of the asymptotes of the curve x'^ — y'' = a''
are y = -\- x and y = — x.
The curve for xy = k is sometimes called the equilateral
hyperbola referred to its asymptotes as axes.
89. Parametric Equations. The parametric equations of the
rectangular hyperbola x^ — y^ = a^ are
X = a sec 8, (1)
and
y = a tan e. (2)
For, dividing (1) and (2) by a, squaring, and subtracting,
-, - ^ = sec" e - tan^ 9 = 1,
which is the same as equation (4) above.
Exercises
1. Find the equations of the following curves after rotation about
the origin through an angle of — 45°.
(o) 2xy = 1; (6) xy = 1; (c) xy = 4; (c) xy = f; (d) xy = 3;
(e) xy - 2 = 0.
2. Show that y^ — x^ = a^ is the equation of the curve 2xy = a'
after rotation about the origin through an angle of + 45°.
3. Show that x^ — y^ = a^ is the equation of the curve 2xy = — o'
after rotation about the origin through an angle of -|- 45°.
4. Find the equations of the curves given in exercise 1 after rota-
tion about the origin through an angle of -|- 45°.
6. Find the equations of the asymptotes for x* — y^ — 2x + 4y = 7.
Hint: Completing the squares
{X - 1)» - (2/ - 2)2 = 4
166 ELEMENTARY MATHEMATICAL ANALYSIS
Since the assnnptotes for i' — ^' — 4 are y •• ± x, the asymp-
totes for
(X - 1)2 - (V - 2)2 = 4 are 2/ - 2 = ± (a; - 1),
or
y = X + 1 and y +x = 3.
6. Find the equations of the asymptotes and sketch the curves for
(a) x' - y' +2x + ^y = 4;
(6) 2x^ - 22/2 + 4x - 81/ = 0.
90. The Hyperbola of Semi-axes a and b. The ellipse was
defined as the curve produced by lengthening or shortening all
ordinates of the circle x^ + y' = a', an amount proportional
to their lengths. Attention has been called to the fact that such
a curve results also from the orthographic projection of the circle,
or from taking the section of a right circular cylinder by a plane.
The parametric equations of the circle are
X = a cos 6,
and
y = a sin d;
and the parametric equations of the ellipse derived from this
circle as described above are
X = a cos d,
and
y = b sin 6.
Let us define the hyperbola as the curve obtained from the
equilateral hyperbola, x^ — y^ = a^, by shortening or lengthening
all ordinates by an amount proportional to their lengths. Its
equation is then obtained by replacing y in
x'^ — y^ = o' (1)
by my. Hence we have for the equation of the hyperbola
x^ — {myy = a^.
a^
By dividing by a^ and replacing — ^ by h'^, we obtam
m
a" b'
.-L--^, (2)
§91] THE ELLIPSE AND HYPERBOLA 167
the symmetrical form of the equation of the hyperbola. It is
easily shown that
X = a sec d, (3)
and
y = b tan e. (4)
are parametric equations of the hyperbola whose Cartesian equa-
tion is given by (2). For from (3) and (4) we obtain
-' -|-, = sec^e - tan^e = 1.
Note that the parametric equations of the equilateral hyper-
bola, X = asecd and y — a tan 6 bear the same relation to the
parametric equations of the hyperbola, that the parametric
equations of the circle bear to the parametric equations of the
ellipse.
It is seen that if the ordinates of the asymptotes to the equilateral
hyperbola are affected in the same way as the ordinates of the
curve itself, i.e., if the asymptotes are considered as part of the
locus transformed, they are still the asymptotes to the hyperbola
after the transformation.
The equations of the asymptotes of the equilateral hyperbola are
y == ± X.
By the transformation they become
my = + X,
a
or, smce ™ ~ ft'
y=±^x, (5)
which are the equations of the asymptotes of
x^ 7/^ _
;^2 ~ p - 1-
91. Construction of the Hyperbola. To construct the hypsr-
bola draw two concentric circles of radii OA = a and OB = 6,
as in Fig. 87. Divide each circumference, by means of bow
168 ELEMENTARY MATHEMATICAL ANALYSIS [§91
dividers, into the same number of convenient intervals. Lay
ofif, on XOX', distances equal to a sec 5 by drawing tangents at
the points of division on the circumference of the a-circle; also lay
off distances equal to 6 tan 6 on the vertical tangent to the 6-
circle by prolonging the radii of the circle through the points of
division of the circumference. Draw horizontal and vertical
lines through the points of division of MN and XX', respectively,
dividing the plane into a large number of rectangles.
J
i
r 1
—Is/
N
T
G
\x
\,
y
yi
\
Y
^
1 / /y^
V^
\
1
t -
^
M
\
\A.'
w
A
D
/
[
/
\
/
\
■A
\
y.
y
f
bVV'
\
"vV
Y
0
^;
Fig. 87.— The hyperbola xP-ja?- - ^fjV = 1.
The point of intersection of the vertical and horizontal line
corresponding to the same value of 5 is a point on the hyperbola.
The curve may be drawn by starting from the points A and A'
and sketching the diagonals of successive rectangles.
In the above construction, there is no reason why the diameter
of the 6-circle may not be greater than that of the o-circle.
The line A A' = 2a is called the transverse axis, the line BB' =
26 is called the conjugate axis, the points A and A' are called the
vertices, and the point 0 is called the center of the hyperbola.
§91] THE ELLIPSE AND HYPERBOLA 169
Solving the equation (2) §90, for y, the equation of the hyperbola
may be written in the useful form
y = + ~ Vx^^T^. (1)
Compare this equation with the equation of the ellipse, (2) §82.
It is easy to show that the vertical distance PG, Fig. 87, of any
point of the curve from the asymptote G'G can be made as small
as we please by moving P outward on the curve away from 0.
Write the equation of the hyperbola in the form
2/1 = V^^^^, (2)
and the equation of the asymptote GG' in the form
2/2 = ^s. (3)
Then
PG? = 2/2 - 2/1 = ^ (x - Va;^ - a-') (4)
Multiply both numerator and denominator in equation (4) by
X + Vx^ - a^.
PC — ~ =^ (5)
0-X + Vx^ - ffl^
Now, as X increases in value without limit the right side of (5)
approaches zero. Whence
PG = Oa,sx = a,
Exercises
1. Write the symmetrical equation of the hyperbola from the
parametric equations x = 5 sec 6, y = 3 tan 8.
2. Find the Cartesian equation of the hyperbola from the relations
X = 7 sec e, 2/ = 10 tan 8. Note that the graphical construction of
the hyperbola holds if 6 > u.
3. What curve is represented by the equation
25 16
170 ELEMENTARY MATHEMATICAL ANALYSIS [§92
4. What curve is represented by the equation y = ^Vs' — o'?
5. Write the equation of a hyperbola having the asymptoteB
y = ±. (3/4) X, and transverse axis = 24.
6. Show that the curves
x^ + 6i
4?/ + 4 = 0
and
(.X + 3)2 -{y + 2Y = l
are the same, and show that each is a hyperbola.
Fig. 88. — Conjugate hyperbolas.
7. What curve is represented by the equations
X = h + as&a e
2/ = A; + 5 tan e?
8. Discuss the curve x' — 8x — 2y' — 12y = 6 .
92. Conjugate Hyperbolas. Let GAJ'JA'G', Fig. 88, be the
hyperbola whose equation is
6^
1.
(1)
Its transverse axis is A A' = 2a and its conjugate axis is BB' = 26.
*92]
THE ELLIPSE AND HYPERBOLA
171
Its asymptotes O'G and J' J are the diagonals, produced, of the
rectangle constructed upon A' A and B'B as sides.
The hyperbola GBJG'B'J', having B'B as transverse axis and
A A' as conjugate axis and G'G and J' J as asymptotes is called the
conjugate of GAJ'JA'G'.
If Y'OY were the Z-axis and if X'OX were the Y-axis the equa-
tion of the hyperbola
GBJG'B'J' would be
0=
1
(2)
By the above supposition we
have interchanged x and y.
Hence, to get the true equa-
tion we must interchange x
and y in equation (2).
Therefore the equation of
the hyperbola conjugate to
x^/a^ — 2/Y62 = 1 is
r
(3)
Fig. 89 shows a family of
pairs of conjugate hyper-
bolas.
Fig. 89. — A family of conjugate
pairs of hyperbolas with common
asymptotes. (An interference pat-
tern made from a glass plate under
compression. From R. Strauble,
"TJeber die Elsticitats-zahlen una
moduln des Glases." Wied. Ann.
Bd. 68, 1899, p. 381.)
Exercises
1. Sketch on the same pair of axes the four following hyperbolas and
their asymptotes:
a)
a;2 - ^2 =
25
(3)
x'
25 ■
-1-
(2)
a;2 _ 2,8 = _ 25
Find the axes of the hyperbola
(4)
i,y --
X2
25 ■
= ±
2/'_
9
1.
2.
iVx«-
64.
3.
Sketch the
curves:
a;2
4
9
= 1
and
4 ~
■'i-
- —
1
172 ELEMENTARY MATHEMATICAL ANALYSIS [§92
4. Sketch the curves:
and
16
r
16
t
9
= 1.
6. Write the equation of the hyperbola conjugate to
2/ = ± f y/x' - 64.
6. Compare the graphs of:
2/ = + f Vx^
3
7. Show that 3x^
Fig. 90. — Diagram for Exercise 13.
64
2/ = + f Vx' - 16
y = ± i Vx^ - 4
2/ = ± i Vx^ - 1
y = ±1 Va;' - 1/16
2/ = + f Vx^ - 0
42/' — 7a; + 52/ + 2 = 0 is a hyperbola. Find
the position of the center and
of the vertices. The vertices
locate the so-called "limiting
Unes" of the hyperbola. Write
the equations of the asymptotes.
8. Show that a;' - 4a;, - ^y^
+ 42/ = 4 is a hyperbola. Find
the coordinates of its center,
the equations of its asymp-
totes, and the equations of its
limiting hnes.
9. Discuss the graphs:
x^ -y^ = 1
and
2/2 -x
■>■ = 1.
= 2, and find the
10. Discuss the graph 16a;2 — y'^ — 40a; —
hmiting lines.
11. Write the equation conjugate to
^ _^ - 1
4 16
12. Write the equation conjugate to
a;' - 2a! - 2/' - 62/ = 24.
13. A difficult problem : Prove that if a circular cylinder be cut by
§92] THE ELLIPSE AND HYPERBOLA 173
a plane at an angle of 45° to the axis of the cylinder, and if then the
surface of the cylinder be unrolled into a flat surface, the curved
boundary of the surface is a sinusoid. Thus if a stove pipe be cut
at an angle of 45° to its axis, and if then the sheet metal be unrolled
into a flat sheet, the bounding curve is a sinusoid.
In Fig. 90 only one quarter of the cylinder is shown. If P be any
point on the section of the cylinder made by the cutting plane, and if
the length of the arc AD be called e and the distance DP be called y,
the problem is to show that y = sin 8, provided the radius of the cylin-
der be called 1. If the angle of cutting be different from 45°, the
equation of the curve is of the form y = bain 6, where b = tan BOC.
F{x, y) = xy sia - (2)
CHAPTER VI
SINGLE AND SIMULTANEOUS EQUATIONS
93. Notation of Functions. It has been pointed out that the
symbols f(,x), F{x), 4>(,x), ^{x), etc., are used to denote functions of
X. Likewise the symbols f{x, y), F(x, y), <f>(x, y),^ix, y), etc., are
used to represent functions of two arguments x and y. For
example, f{x, y) in a particular problem may be used to stand
xy
for the function / , , „■ . We may indicate this fact by writing
V a;2 +y^
Again we may abbreviate the function of x and y, xy sin -> by
the symbol F(x, y). This abbreviation can be indicated by writing
The equation
F{x, 2/) = 0 (3)
indicates that y is a function of a;; y is a function of x expressed
implicitly. If equation (3) were solved for y giving
V^Kx), (4)
J/ is a function of x expressed explicitly. Equations (3) and
(4) represent the same functional relation between x and y.
Thus x^ + 2/2 — a'' = 0 shows that y is & function of x but the
functional relation is expressed implicitly. If the equation be
solved for y, giving y = ± ^a^ — x"^, the same functional relation
between x and y holds, but now 2/ is an explicit function of x.
In the same problem or discussion the symbols f{x), f(y), f{u),
or /(t)) denote the same functional form although the ftrguments
may differ. If
^(") = V#+t'
174
§94] SINGLE AND SIMULTANEOUS EQUATIONS 175
f{v) means the same function but with every x replaced by y, thus
m - ^
Again, if in a particular problem or discussion
f{x) =x^ + 2x-l,
then fiy) = y^ + 2y - 1,
/(2) = 2= + 2-2 - 1 = 7,
f(- 1) = (- 1)^ + 2(- 1) - 1 = - 2,
/(O) = 0 + 0 - 1 = - 1.
Exercises
1. URx) mx' + 3x+2, find/(2/);/(3);/tO);/(-l);and/(-2).
2. If /(a;) ^ s' + 2x^ + X, find A-1); /(O); /(+1J; f(z); Al/v);
and /(<2).
3. liF(e) =sinfl, findf(,r/2);Ftx);F(0);F(x/6);F(V3)andF(|j-)
4. If*>(fl) s tanS, find¥>(0);¥>(7r/6);«>lir/3);#>(ir'/2);¥'(7r)and*>(|ir).
5. If /(I, 2/) ^ -==^' find/(2, 1);/ (0, 2);/ fe «) and/(m, n).
V a;'' + y'
Hint: To find /(2, 1) replace x by 2, and ?/ by 1 in the given func-
tion of X and y.
94. A poljmomial in x of the nth degree is defined as
aox" + aia;""' + UiX"'^ + + a„_ia; + On,
where the symbols, ao, ai, 02, . . ., stand for any real con-
stants whatsoever, positive or negative, integral or fractional,
rational or irrational, and where n is any positive integer. If
none, of the coefficients are zero the number of terms in a
polynomial in x of the nth degree is (n + 1).
In what follows in this chapter /(re) is supposed to stand for a
polynomial in x.
96. The Remainder Theorem. Let
f{x) = aax" + Oia;"-^ + ajx"-^ H- . . -|- On-iX + On. (1)
Then /(r) =aor» + aif-^ -|- a^r''-'^ + -f (h-\r + a„. (2)
By subtracting (2) from (1),
/(*) ~ /('■) — "oCa;" — r") + a\{x''~^ — r"~^) -f
+ o^-i(s - r). (3)
176 ELEMENTARY MATHEMATICAL ANALYSIS
The right-band side of this equation is made up of a series of terms
containing differences of like powers of x and r, and, hence, by the
well-known theorem in factoring,' each binomial term is exactly
divisible by {x — r). The quotient of the right-hand side of (3)
by {x — r) may be written out at length, but it is sufficient to
abbreviate it by the symbol Q{x) and write
fix) -fir)
or
:/^=QW+;^- (5)
Equation (5) shows that f(r) is the remainder when f(x) is
divided by (a; — r). Thus we have the Remainder Theorem :
If a polynomial in x be divided by (a; — r), the remainder which
does not contain x is obtained by writing, in the given function, r in
place of x. This theorem shows, for example, that the remainder
of the division
(.t' - 6x2 + iix _ 6) -f. (x - 4) is 43 - 6(4)2 + n(4) - 6, or 6;
also that the remainder of the division
(x' - 6x2 + iia; _ 6) 4- (x -I- 1)
is
(- 1)3 - 6(- 1)2 -1^ 11(- 1) -■ 6 = - 24.
The theorem enables one to write the remainder without actually
performing the division.
Exercises
Without performing the division find the remainder of the following
divisions :
1. (x2 + 3x - 2) -=- (x - 1).
2. (x' + 3x2 + 2x -1) ^ (x - 2).
3. {x* + 4x= + 3x2 _ 62; - 1) -^ (I + i).
4. (x' - 3x2 + 2x - 1) -^ (x -2).
6. (x2 + 3x + 2) + (x -h 1).
6. (x2 + 3x + 2) -i- {x + 2).
96. Factor Theorem. From equation (5) of the preceding
section, we see that if fir) is zero, the remainder of the division of
> See Appendix, Chapter XV, p. 159.
§96] SINGLE AND SIMULTANEOUS EQUATIONS 177
fix) by (x — r) is zero, or /(a;) is exactly divisible by (x — r), i.e.
(x — r) is a factor of f(x). Thus we have the Factor Theorem:
If a polynomial in x becomes zero when r is written in the place of x,
{x — r) is a factor of the polynomial. This means, for example, that
if 3 be substituted for x in the function x' — 6a;^ + Ha; — 6 and
if the result 3^ - 6(3)^ + 11(3) - 6 is zero, then {x - 3) is a
factor of x^ — Qx'^ + lis — 6.
The value r of the argument x that causes the function to take
on the value zero has already been named a root or a zero of the
function. ' The factor theorem may, therefore, be stated in the
form : A polynomial in x is exactly divisible by (x — r) where r is any
root of the polynomial.
The familiar method of solving a quadratic equation by
factoring is nothing but a special case of the present theorem.
Thus, if
x^ - 5x + Q = 0,
(x - 2)(x - 3) = 0;
and the roots are x = 2 and x = 3. The numbers 2 and 3 are
such that when substituted in x^ — 5x + 6 the expression is
zero; and the factors of the expression are x — 2 and ic — 3 by
the factor theorem.
Exercises
1. Tabulating the cubic polynomial /(a;) = x' — 6a;* -|" H^ —6, we
obtain:
X -3 - 2 - 1 - 0 1 1.5 2 2.5 3 4
fix), -120, -60, -24, -6, 0, +0.375, 0, - 0.375, 0, 6
What is the remainder when the function is divided by .r — 4?
By I + 2? By a; + 3? By a; - 1.5? By a: - 3?
Name three factors of the above function.
2. Find the remainder when a;* — 5x^ + 12.1;^ + 4a; — 8 is divided
by a; - 2.
3. Show by the remainder theorem that x" + a" is divisible by
X + a when n is an odd integer, but that the remainder is 2o" when n
is an even int^er.
4. Without actual division, show that x* — ix' — 7x — 24 is
divisible by a; — 3.
6. Show that a* + a^ — ab' — ¥ is divisible by o — b.
12
178 ELEMENTARY MATHEMATICAL ANALYSIS [§97
6. Show that (x + l)Ha; - 2) - 4(a; - l)(a: - 5) + 4 is divisible
by I - 1.
7. Show that &x^ - 3x* - 5a;' + 5a;' - 2x - 3 is divisible by
x + 1.
8. Show that (b - c){h + c)' + (c - a){c + a)' + (o - b)(a + b)'
is divisible by (5 — c)(_c — a) {a — b).
Hint: First consider the function as a polynomial in 6; then as a
polynomial in c; and then as a polynomial in a.
9. Show that (6 - c)' + (c - o)' + (o - 6)' is divisible by
(6— c){,c — a)(fl — b).
97. An Equation with Given Roots. The factor theorem enables
us to build up a polynomial having given roots. If, for example, 1,
2, and 3 are roots, 2; — 1, x — 2, and a; — 3, are factors of the poly-
nomial. Hence (x — l)(x — 2) (x — 3), or x' — 6x' + llx — 6
is a factor of the polynomial. Introducing another factor k, which
does not contain x, cannot introduce another root, as a, for k can-
not contain the factor (x — a).
For the same reason, multiplying the equation x' — 6x^ -)- llx
— 6 = 0 by fc, when k does not contain x, cannot introduce roots,
or solutions, in the equation. On the other hand if the equation
be multiplied by a function of x, roots of the equation may be
introduced or removed. For, clearly, if the multiplier contains the
factor (x — a), the root a will be introduced; and if the multiplier
contains the factor (x — 1) in its denominator, the factor (x — 1)
will be divided out from both numerator and denominator, if it
is a factor of the numerator and the root 1 wiU be removed from
the function.
Exercises
Build up polynomial equations having the following numbers for
roots:
1. 1, 3, and 4. 2. -1, 2, and -3. 3. 0, 2, and -1. 4. 1, 0, 0,
and 2.
98. Legitimate and Questionable Transformations. If one
equation is derived from another by an operation which has no
effect one way or another on the solution, it is spoken of as a
legitimate transformation ; if the operation is of such a nature that
it may ha.ve an effect upon the roots, it is called a questionable
§98] SINGLE AND SIMULTANEOUS EQUATIONS 179
transfonnation, meaning thereby that the effect of the operation
requires examination.
In performing operations on the members of equations, the
effect on the solution must be noted, and proper allowance
made in the result. It cannot be too strongly emphasized that
the test for any solution of an equation is that it satisfy the original
equation. "No matter how elaborate or ingenious the process
by which the solution has been obtained, if it do not stand this
test it is no solution; and, on the other hand, no matter how simply
obtained, provided it do stand this test, it is a solution."^
By the principles or axioms of algebra, an equation remains
true if we unite the same number to both sides by addition or
subtraction; or if we multiply or divide both members by the
same number, not zero; or if like powers or roots of both members
be taken. But we have indicated in the preceding section that
these operations may affect the number of roots of the equation.
This is obvious enoHgh in the case already cited. Sometimes,
however, the operation that removes or introduces roots is so
natural and its effect is so disguised that the student is apt not to
take due account of its effect. Thus, the roots of
3(x - 5) = x(x - 5) + x' - 25 (1)
are — 1 and 5, for either of these when substituted for x will
satisfy the equation. Dividing the equation through by a; —5,
the resulting equation is
3 = a; + a; + 5.
This equation is not satisfied by a; = 5. One root has disappeared
in the transformation. It is easy to keep account of this if (1)
be given in the form
(a; - 5)(a; + 1) = 0,
but the fact that a factor has been removed may be overlooked
when the equation is written in the form first given.
A very important effect upon the roots of an equation results
from squaring both members. The student must always take
proper account of the effect of this common operation. To il-
lustrate, take the equation
a; + 5 = 1 - 2a;. (2)
' Chrystftl's Algebra,
180 ELEMENTARY MATHEMATICAL ANALYSIS
It is satisfied only by the value a; = — f . Now, by squaring
both sides of the equation, we obtain
a;'' + lOs + 25 = 1 - 4x + ix',
which is satisfied by either a; = 6 or a; = — |. Here obviously,
an extraneous solution has been introduced by the operation of
squaring both members.
It is easy to show that squaring both members of an equation
is equivalent to multiplying both sides by the sum of the left and
right members. Thus, let any equation be represented by
L(x) = R^x) (3)
in which L(x) represents the given function of x that stands on
the left-hand side of the equation and R{x) represents the given
function of x that stands on the right-hand side of the equation.
Squaring both sides,
[Lix)]^ = [B(,x)V.
Transposing,
[L(,x)r - [R\x)V = 0,
factoring,
[Lix) + R(x)] mx) - Rix)] = 0.
But (3) may be written
L{x) - R{x) = 0.
Thus, by squaring the members of equation (3) the factor
L(x) + R(x) has been introduced.
The sum of the left- and right-hand members of (2), above, is
6 — a;. Hence, squaring both sides of (2) is equivalent to the
introduction of this factor, or thq operation introduces the root
6, as already noted.
As another example, suppose that it is required to solve
sin a cos a = \ (4)
for a < 90°. Substituting for cos a, Equation (4) becomes
sin aVl — sin" a. - \, (5)
squaring
sin^ a(l — sin'' a) = y^,
§98] SINGLE AND SIMULTANEOUS EQUATIONS 181
completing the square
sin* a. — sin'' a-\-\ = j-^-.
Hence,
sin a = + Vi + i \/3
=• ± 0.9659 or ± 0.2588.
Only the positive values satisfy (4); the negative values were
introduced in squaring (5). If, however, the restriction a < 90°
be removed, so that the radical in (5) must be written with the double
sign, then no new solutions are introduced by squaring.
Among the common operations that have no effect on the solu-
tion are multiplication or division by known numbers, or addition
or subtraction of like terms to both members; none of these intro-
duce factors containing the unknown number. Taking the
square root of both numbers is legitimate if the double sign be
given to the radical. Clearing of fractions is legitimate if it be done
so as not to introduce a new factor. If the fractions are not in
their lowest terms, or if the equation be multiplied through by an
expression having more factors than the least common multiple
of the denominators, new solutions may appear, for extra factors
are probably thereby introduced. Hence, in clearing of fractions,
the multiplier should be the least common denominator and the
fractions should be in their lowest terms. This, however, does not
constitute a sufficient condition, therefore iAe only certainty lies
in checking all results.
Exercises
Suggestions: It is important to know that any equation of the
form
oa;2» + bx'' + c = 0
can be solved as a quadratic by finding the two values of a;". Fre-
quently equations of this type appear in the form
dx' + ex~^ = f.
Likewise any equation of the form
aj(x) + 6V7(S) + c = 0
can be solved as a quadratic by finding the two values of VjCi) and
182 ELEMENTARY MATHEMATICAL ANALYSIS
then solving the two equationa resulting from putting ■\/f(x) equal to
each of them. One of these usually gives extraneous solutions.
These two tjrpes occur in the exercises given below.
Since operations which introduce extraneous solutions are often
used in solving equations, the only sure test for the solution of any
equation is to check the results by substituting them in the original
equation.
Take account of all questionable operations in solving the following
equations:
3a; 6,9
+ 7.- Note : 3 la not a, root.
' X -3 a; + 3a;-3
2. {x^ + 5x + 6)/{x - 3) + 4a; - 7 = - 15.
3. 3(a; - 5){x - l){x - 2) = (x - 5)(x + 2)(.x + 3).
Note : Divide by (a; — 5), but take account of its effect.
4. x'/a + ax = x''/h + 6a;.
6. oa;(ca; — 36) = 5o(36 — ex).
6. a;' — Ji* = n — a;.
7. (a; - 4)» + (a; - 5)» = 31[(a; - 4)^ - (a; - 5)'^]. Divide by
(a; - 4) + (a; - 5) or 2x - 9.
a;^ — 3a; 1
8. ■ _ . — h 2 + _ - =0. If the fractions be added, multi-
plication is unnecessary. There is only one root.
9. X = 1 - Va;' - 7.
10. Va: + 20 - Va; - 1 - 3 = 0.
11. \/l5/4 + a; = 3/2 + -y/g.
12. 20a;/ V 10a; - 9 - VlOx - 9 = IS/VlOa; -9+9.
13. — ;= , = ;;. Consider as a proportion and take
y/x- y/x-Z ^-^
by composition and division.
14. a;^_+ 5/2 = (13/4)x>^.
16. y/x^ - 2y/x + a; = 0. Divide by y/~x.
16. 2V'a;2 -5x + 2 - x' + 8x = 3x - 6. Call a;^ - 5a; + 2 = u'.
17. 4a;2 - 4a; + 20\/2a;2 - 5a; + 6 = 6a; + 66.
18. x-^ - 2a;-i = 8. 22. 8x^ - Sx'^ = 63.
19. x^^ - 5a;^i +4 = 0. 23. (x - a)" - 3(x - a)-» = 2.
20. 110a;-* + 1 = 21a;-2 24. 2a;^ - 3a;^ + x = 0.
21. Vx + 4x-J^ = 5.
99. Intersection of Loci. In §41 it was shown that the coordi-
nate of the points of intersection of two loci could be found by
solving the equations of the loci considered as simultaneous
equations.
SINGLE AND SIMULTANEOUS EQUATIONS 183
Let all terms of an equation be transposed to the left-hand
member, rendering the right-hand member zero. Let this left-
hand member be abbreviated by u. The equation then takes the
form
M = 0. (1)
In a similar way let a second equation be put in the form
« = 0.
(2)
Fig.
91 . — Intersections
curves.
Let the graphs for equations (I) and (2) be represented in Fig.
91. The coordinates of any point on curve (1) make u equal to
zero. The coordinates of any point
on curve (2) make v equal to zero.
Consider the graph of
u + kv = Q, (3)
where h is any constant. The co-
ordinates of a point of intersection
of the u and v curve satisfy equa-
tion (3). For, these coordinates
make u zero and they make v zero,
then they make u + kv zero.. Fur-
ther, the coordinates of a point on
the u curve which is not on the v
curve do not satisfy equation (3). For these coordinates make
u zero but do not make v zero, then they do not make u + kv
zero. Similarly the coordinates of a point on the v curve which
is not on the u curve do not satisfy equation (3). Hence the
graph of (3) passes through all points of intersection of the
u and V curves but does not intersect these curves in any other
points. Thus to find the coordinates of the points of intersec-
tion of the u and v curve we may solve (1) and (3) or (2) and (3)
as simultaneous equations.
The locus of the equation
WW = 0 (4)
is the M and v curves considered as a single locus. For, the coordi-
nates of a point on the u curve make u zero, then they make uv
zero. Similarly the coordinates of a point on the v curve make
184 ELEMENTARY MATHEMATICAL ANALYSIS
uv zero. The coordinates of a point neither upon the u curve
nor upon the v curve make neither u nor v zero, then they cannot
make uv zero. Hence the locus of (4) consists of all points on
the u and v curve but of no other points.
To find the points of intersection of the circle x^ + y' = 25 and the
straight line x + y = 7 yre solve the equations by the usual method, as
follows:
x^ + y' = 25\ (5)
X +y = 7j (6)
The graphs are a circle and a straight line, as shown in (1), Fig. 92.
Squaring the second equation, the system becomes
x^+y'> = 25\ (7)
X' + 2xy + 2,2 = 49 / (8)
The second equation represents the two straight lines shown in (2)
Fig. 92. The effect of squaring has been to introduce two extraneous
solutions corresponding to the points Ps and Pt. For, eCiuation (8)
may be written {x + y + 7)ix + y — 7) =0 while (6) from which it
was derived is x + y — 7 = 0.
Multiplying (7) by 2 and subtracting (8) from it, the last pair of
equations becomes
x^ -2xy + y* = l\ (6)
x' + 2xy + v" = 49 J (7)
which gives the four straight lines of Fig. 92, (.4). Taking t^e square
root of each member, but discarding the equation x + y + 7 = 0,
because it corresponds to the extraneous solutions introduced by the
questionable operation, we have:
x-y = ±l\ (8)
(9)
-2/= ±1\
+ y =7 /
By addition and subtraction we obtain the results:
(10)
X = S\
y=4J
a; =4"!
2/ = 3/
(11)
represented by the intersections of the lines parallel to the axes shown
in Fig. 92, (5).
§99] SINGLE AND SIMULTANEOUS EQUATIONS 185
This is a good illustration of the graphical changes that take place
during the solution of simultaneous equations of the second degree.
The ordinary algebraic solution consists, geometrically, in the succes-
sive replacement of loci by others of an entirely different kind, but all
passing through the points of intersection (as Pi, Pa, Fig. 92) of the
original loci. The final locii are straight lines parallel to the axes.
FiQ. 92. — Graphic representation of the steps in the solution of a
certain set of simultaneous equations.
Exercises
Find the coordinates of the points of intersection of the following
pairs of equations; sketch curves representing all equations involved
in the solution:
1. xy = 1
3x - 5y = 2
2. X' + y' = 5
y^ = 4x
Hint fob Ex. 4: Let u = x^andw
3. x' + X = 4j/*
3j + 6j/ = 1
4. x2 + 2/2 = 9
a;2 - yi = 4
= 1/2. Solve for u and v.
186 ELEMENTARY MATHEMATICAL ANALYSIS [§100
Solve graphically the following :
6. x' + y' = 25 6. x' + y^ = 25
X +y =2 x^ +y^ + 2x -6y + 6 =0
7. y = x^ + X — I
xy = 1.
100. Quadratic Systems.' Any linear-quadratic systena of
simultaneous equations, such as
y = mx + k
ax'' + hy^ + 2hxy -\- 2gx + 2fy + c = 0
can always be solved analytically; for y may readily be eliminated
by substituting from the first equation into the second. A
system of two quadratic equations may, however, lead, after
elimination, to an equation of the third or fourth degree; and,
hence, such equations cannot, in general, be solved until the
solutions of the cubic and bi-quadratic equations are known.
A single illustration will show that an equation of the fourth
degree may result from the elimination of an unknown number
between two quadratics. Thus, let
x^ — y = 5x
a;2 + j/2 = 10.
From the first, y = x^ — 5x. Substituting this value of y in the
second equation, and performing the indicated operations, we
obtain
a;4 _ lOx' + 26s^ - 10 = 0.
WhUe, in general, a bi-quadratic equation results from the
process of elimination from two quadratic equations, there are
special cases of some importance in which the resulting equation
is either a quadratic equation or a higher equation in the quadratic
form. Two of these cases are:
(1) Systems in which the terms containing the unknown num-
bers are homogeneous; that is, systems in which the terms con-
^ A large part of the remainder of this chapter can be omitted if the students
have had a good course in algebra in the secondary school.
§101] SINGLE AND SIMULTANEOUS EQUATIONS 187
taining the unknown numbers are all of the second degree with
respect to the unknown numbers, such, for example, as
x'' — 2xy = 5
3x^ - lOy^ = 35.
(2) Systems in which both equations are symmetrical; that is,
such that interchanging x and y in every term does not alter the
equations; for example
x' + y^ - X - y = 78
xy + X + y = 39.
101. Unknown Terms Homogeneous. The following work
illustrates the reasoning that will lead to a solution when applied
to any quadratic system all of whose terms containing x and y
are of the second degree. Let the system be
x^ — xy = 2
2x^ + 2/2 = 9. (1)
Divide each equation by x'^ (or y'^), then
1 - iy/x) = 2/x^
2 + (y/xr = Vx'. (2)
Since the left members were homogeneous, dividing by x' renders
them functions of the ratio (y/x) alone; call this ratio m. Then
equations (2) contain only the unknown numbers m and x^.
The latter is readily eliminated by subtraction, leaving a quad-
ratic for the determination of m. When m is known, substituting
in (2) determines x, and the relation y = mx determines the
corresponding values of y.
The above illustrates the principles on which the solution is based.
In practice, it is usual to substitute y = mx at once, and then eliminate
x' by comparison; thus, from the substitution y = mxin (1), we obtain
x' - mx^ = 2
2x' + mV = 9. (3)
188 ELEMENTARY MATHEMATICAL ANALYSIS [§101
Thence,
Whence,
or
a;2 = 2/(1 - m)
x^ = 9/(2 + m').
2/(1 -m) = 9/(2 + m^),
2m' + 9m = 5.
\
\
Y
\
\
\
\
4
/
V
y//
\
3
//
u\
2 \^
1 y/
X' / J
' JC
-4 -3 - 2 /-I ^^Z'
%
2 3 4
/
//
A
- a = - Vs Vi"
//
•4 \
\
'
\
\
Y
1
(4)
(5)
(6)
Fig. 93. — Solutions of a set of simultaneous quadratics given graph-
ically by the coordinates of the points of intersection of an ellipse and
hyperbola.
Factoring,
whence.
Hence,
(2m - l)(m 4- 5) = 0
m = 1/2 or — 5.
X = + 2 or + (l/3)-v/3
2/ = + 1 or + (5/3)v'3.
(7)
(8)
(9)
§102] SINGLE AND SIMULTANEOUS EQUATIONS 189
These solutions should be written as corresponding pairs of values as
follows:
X = 2 X. = -2 X = (1/3)V3 a; = - (l/3)\/3
y = l v=-l t/='-(5/3)V3 y= (5/3)^3
This system can readily be solved without the use of the mx sub-
stitution by merely solving the first equation fpr y and substituting
in the second.
Graphically (See Fig. 93), the above problem is equivalent to
finding the intersections of the curves :
x(x - y) = 2
(V2x)' + y' = 9
The first is a curve with the two asymptotes x = 0 and x — y = 0.
That these lines are asymptotes is readily seen if the equation be
2
put in the form y = x If a; is positive, y is less than x, or the
curve is below the line y = x. If x is negative y is greater than x, or
the curve-is above the line y = x. As x increases in numerical value,
2
- approaches zero and the curve approaches the line y = x. As a;
approaches zero, y increases without limit. As a matter of fact, the
curve is a hyperbola, although proof that such is the case cannot be
given until the method of rotating any curve about the origin has been
explained. The second curve is obviously an ellipse generated from
a circle of radius 3 by shortening the abscissas in the ratio ■y/2 : 1. The
two curves intersect at the points:
X =2 - 2 0.557 . . . -0.557 ...
2/ = 1 - 1 - 2.887 ... +2.887 ...
The auxiliary lines, y = ^x and y = — 5x, made use of in the solution
are shown by the dotted lines.
102. Symmetrical Systems. Simultaneous quadratics of this
type are readily solved analytically by solving for the values of
the binomials x -\- y and x — y. The ingenuity of the student
will usually, show many short cuts or special expedients adapted
to the particular problem. The following worked examples point
oat some of the more common artifices used.
1. Solve
x + y =Q (1)
xy = 5. (2)
190 ELEMENTARY MATHEMATICAL ANALYSIS [§102
Squaring (1)
x^ + 2xy + y^ = 36. (3)
Subtracting four times (2) from (3)
x'- - 2xy + 2/= = 16.
Whence
But from (1)
Therefore
a; = 5
2/ = l
2. Solve
X -y =
X +y =
and
X2 + 2/2 :
± 4.
6.
= 34
a; = 1
2/ =5.
(1)
xy ■■
Adding two times (2) to (1)
= 15
(2)
x'
> + 2xy +
y' = 64.
(3)
Subtracting two times (2) from (1)
X
Whence, from (3) and (4)
2 - 2xy +
X +y =
2/2=4-
±8
(4)
Therefore
X - V =
±2.
X = 5 X =3
a; = - 5
' X =
-3
^ = 3 y = 5
y = -3
y =
-S
The hyperbola and circle j
by the student.
3. Solve
represented by (1) and (2) should be drawn
x> + y^ = 72 (1)
X +y =
Cubing (2)
a;3 4- 3a;2j, ^ ^xy'
= 6.
' + y' = 216.
(2)
(3)
Subtracting (1) and dividing by 3
4
whence, since
xy(x + y)
X +y
= 48,
= 6
(4)
we have
y
= 8.
(5)
§102] SINGLE AND SIMULTANEOUS EQUATIONS 191
From (2) and (5) proceed as in example 1, and find
1 = 4 , X = 2.
r> and ,
y = 2 2/ = 4
Otherwise, divide (1) by (2) and proceed by the usual method.
4. Solve
a;2 + SI/ = ^ (a; + J/) (1)
y-'-'rxy = ^- (x + v). (2)
Adding (1) and (2)
{x + yy - 6{x + y) '= 0, (3)
whence,
X + 2/ = 0 or 6. (4)
Now, because x + y is a factor of both members of (1) and (2), the
original equations are satisfied by the unlimited number of pairs of
values of x and y whose sum is zero, namely, the coordinates of all
points on the line x -\- y = Q.
Dividing (1) by (2), we get
x/y = 7/11.
This, and the line x -\- y = &, from (4), give the solution:
y = T/Z
y = 11/3.
Graphically, the equation (1) is the two straight lines i;
{x-7/3){x + y) =0.
Equation (2) is the two straight lines
(2/ - n/3){x+y) =0.
These loci intersect in the point (7/3, 11/3) and also intersect every-
where on the line x + y = 0.
Exercises
1. Show that
3-2 + J/2 = 25
X + y = 1
has a solution, but that there is no real solution of the system
a;2 + j/2 = 25
X +y = U.
192 ELEMENTARY MATHEMATICAL ANALYSIS [§103
2. Do the curves
Do the curves
3. Solve
x' + y' = 25
xy — 100, intersect?
a;2 + 2/2 = 25
xy = 12, intersect?
(x^ + y'){x + y) = 272
x^ + y^ + x'+ y = 42.
Note : Call x^ + y^ = u, and x + y = v.
4. Show that there are four real solutions to
x^ -\- y^ - \2 = X -\-y
xy-\-S = 2{x +y).
5. Solve x'' -\- y^ -\- x + y = li
xy = 6.
103. Graphical Solution of the Cubic Equation. The roots of a
cubic x' + ax' + j3x + 7 = 0 (where a, /3, and 7 are given known
numbers) may be determined graphically as explained in §40.
Another method of solving the cubic equation graphically
will now be given. The roots of the equation
x^ + ax^ + ^x + y = 0 (1)
are the JST-intercepts for the graph of
y = x^ + ax' + ^x + y. (2)
If we replace x in equation (2) by (x — k), where fc is a constant,
the equation (2) becomes
y = ix-ky+ a(.x - ky + /3(a; - k) + y,
or
y = x^ + {a- Zh)x' + (j3 - 2ak + ^k')x
-{¥ - al<!,'+ fik -y). (3)
a
It will be noticed that if k is chosen equal to -5 the coefficient of
x^ vanishes and equation (3) becomes
y = x^ + ax + b, (4)
§103] SINGLE AND SIMULTANEOUS EQUATIONS 193
when a stands for the coefficient of x and h stands for the absolute
term of equation (3).
Since the graph for (4) differs from the graph of (2) only in that
it is translated -s units parallel to the Z-axis, the X-intercepts for
the first graph are ^ units greater (less if a is negative) than the
X-intercepts for the second graph. Hence, if the roots of
x^ + ax + h (5)
can be found, these roots decreased by o are the roots of (1).
Since an equation of the form
(2) can always be put in the form
of equation (4), we shall only
consider cubic equations of the
form
x^ + ax + h = Q. (6)
Consider the system of curves
2/ = x= (7)
y = — ax — b. (8)
Equation (7) gives the cubic pj^ 9 4.— Construction for
parabola, and (8) the straight graphical solution of .t' + a x+
line. Fig. 94. ^ = "•
Let P be a point of intersec-
tion of the cubic and the straight line. Let OD be the abscissa
of the point P. The value of OD is a root of equation (6). For
OD is a value of x for which a;' = — ax — b, or for which a;' +
ax + b = 0.
In drawing the graph of the cubic parabola, it is desirable to use, for
the ^-seale, one-tenth of the unit used for the x-scale, so as to bring a
greater range of values for y upon an ordinary sheet of coordinate
paper. The cubic parabola graphed to this scale is shown in Kg. 95.
The diagram gives the solution of s' — a; — 1 = 0. The graphs
y = x' and y = x + 1 aie seen to intersect at x = 1.32. This, then,
should be one root of the cubic correct to two decimal places. The
line y = X + 1 cuts the cubic parabola in but one point, which shows
that there is but one real root of the cubic. To obtain the imaginary
194 ELEMENTARY MATHEMATICAL ANALYSIS [§103
Fia. 95. — Graphic solution of the cubic a;^ — a; — 1 = 0 and
s3 _ 10 a; - 10 = 0.
§103] SINGLE AND SIMULTANEOUS EQUATIONS 195
roots, divide a;^ — a; — 1 by x — 1.32. The result of the division,
retaining but two places of decimals in the coefficients, is
s2 + 1.32x + 0.7424.
. Putting this equal to zero and solving by completing the square, we
find
X = - 0.66 + 0.55V - 1,
in which, of course, the coefficients are not correct to more than two
places.
The equation
x^ - lOx - 10 = 0 (9)
illustrates a case in which the cubic has three real roots. The straight
line y = IQx + 10 cuts the cubic parabola (See Fig. 95) at x = — 1.2,
X = — 2.4, and x = 3.6. These, then, are the approximate roots.
The product
(x + 1.2) (x + 2.4) (x - 3.6) = x' - lO.OSx - 10.37
should give the original equation (9). This result checks the work
to about two decimal places.
The x-scale of Fig. 95 extends only from — 6 to + 5. The same
diagram may, however, be used for any range of values by suitably
changing the unit of measure on the two scales; thus, the divisions of
the x-scale may be marked with numbers 5-fold the present numbers,
in which case the numbers on the y-aaaXe must be marked with num-
bers 125 times as great as the present numbers. These results are
shown by the auxiliary numbers attached to the i/-scale in Fig. OS.'
It is obvious that a similar process will apply to any equation of the
form
X" + ox + 6 = 0.
Exercises
Solve graphically the following equations and check each result
separately :
1. x' - 4x -I- 10 = 0 4. x' - 15x - 5 = 0.
2. x' - 12x - 8 = 0. 5. x' - 3x + 1 = 0.
3. x' -I- X - 3 = 0. 6. x^ - 4x - 2 =0.
7. 2 sin 9 -f- 3 cos e = 3.5.
» For other graphical methods of solution of equationSt see Runge's Graphical
Methods," Columbia University Press, 1912. More work on the graphical solution
of the cubic will be found in Schultze, " Advanced Algebra," p. 484.
196 ELEMENTARY MATHEMATICAL ANALYSIS [§104
Note: Construct on polar paper the circles p = 3.5 and p =
2 sin 9 + 3 cos 0.
8. 2x + sin a; = 0.6.
Note: Find the intersec Jon oiy = sin x and the line y= — 2a; + 0.6.
If 1.15 inches is the amplitude of 3/'= sins;, then 1.15 inches must be the
unit of measure used for the construction of the line y = — 2a;+ 0.6.
9. x' + X + 1 + 1/x = 0.
10. Show that a;' + a.T + b = 0 can have but one real root if a > 0.
104. Method of Successive Approxunations. It must be re-
membered that the graphic methods of solving numerical equa-
tions by finding one or both coordiaates of points of intersection of
graphs, gives results only approximately correct. The degree of
accurately depends upon the scale of the drawing and upon
the accuracy with which the graphs are constructed. The results
thus obtained may be used as a first approximation to the solution
by a method illustrated below
Suppose that it is required to find to four decimal places one root of
x' — X — 1 = 0. See §103 and Fig. 95. The graphic method gives
X = 1.32. This is the first approximation. A second approximation
is found as follows :
Substituting 1.32 for x in
y = x' — X — \ J (1)
gives — 0.0200 for y. This shows that 1.32 is not the exact value for
y. Substituting 1.33 for a; gives 0.0226 for y. Put these results in
tabular form
X
y
p
Q
1.32
1.33
-0.0200
+0.0226
Differences
0.01
0.0426
This shows that the X-intercept of the graph of the given
equation is between the points P and Q, Fig. 96. Thus a root of
x' — a; — 1 is greater than 1.32 and less than 1.33. Now reason as
follows: The actual root lies between 1.32 and 1.33, and the zero value
of y corresponds to it. This zero is approximately 200/426 of the way
between the two values of y. Hence if the curve be nearly straight
between x = 1.32, and x = 1.33, the desired value of x is approxi-
mately 200/426 of the way between 1.32 and 1.33 or it is x = 1.3247
approximately. This value is probably correct to the fourth decimal
§104] SINGLE AND SIMULTANEOUS EQUATIONS 197
place. The next step will show that this result is correct to four
decimal places.
To find a third approximation we build another table of values:
1.3247
1.3248
Differences 0.0001
y
-0.0000766
+0.0003499
0.0004265
Fig. 96. — Method of approximation to a root of an equation.
Reasoning as before, we get x = 1.324718 which is very likely true
to the last decimal place.
The above method is applicable to an equation like exercise 8
above. In fact it is the only numerical method that is applicable tn
such cases.
Exercises
Find correct to four decimal places the roots of:
1. x' -ix + 10 = 0.
2. X' - 12x -8=0. See Exercises 1 and 2, §103.
CHAPTER VII
PERMUTATIONS AND COMBINATIONS;
THE BINOMIAL THEOREM ]
105. Ftmdamental Principle. If one thing can be done in n
different ways and another thing can be done in r different ways,
then both things can be done together, or in succession, in n Xr
different ways. This simple theorem is fundamental to the work
of this chapter. To illustrate, if there be 3 ways of going from
Madison to Chicago and 7 ways of going from Chicago to New
York, then there are 21 ways of going from Madison to New York.
To prove the general theorem, note that if there be only one
way of doing the first thing, that way could be associated with
each of the r ways of doing the second thing, making r ways
of doing both. That is, for each way of doing the first, there are
r ways of doing both things; hence, for n ways of doing the first
there are n X r ways of doing both.
Illustrations: A penny may fall in 2 ways; a common die may
fall in 6 ways; the two may fall together in 12 ways.
In a society, any one of 9 seniors is eligible for president and any one
of 14 juniors is eligible for vice-president. The number of tickets
possible is, therefore, 9 X 14 or 126.
I can purchase a present at any one of 4 shops. I can give it away
to any one of 7 people. I can, therefore, purchase and give it away in
any one of 28 different ways.
A product of two factors is to be made by selecting the first factor
from the numbers a, b, c, and then selecting the second factor from the
numbers x, y, z, u, v. The number of possible products is, therefore, 15.
If a first thing can be done in n different ways, a second in r
different ways, and a third in s different ways, the three things
can^be done in n X r X s different ways. This follows at once
from the fundamental principle, since we may regard the first
198
§106] PERMUTATIONS AND COMBINATIONS 199
two things as constituting a single thing that can be done in nr
ways, and then associate it with the third, making nr X s ways
of doing the two things, consisting of the first two and the third.
In the same way, if one thing can be done in n different ways, a
second in r different ways, a third in s, a fourth in t, etc., then all
can be done together inn X r X s X t "different ways.
Thus, n different presents can be given to x men and a women
in (x + a)" different ways. For the first of the n presents can
be given away in (x + a) diiferent ways, the second can be given
away in (x + a) different ways, and the third in (a; + a) different
ways and so on. Hence, the number of possible ways of giving
away the n presents to {x + a) men and women is
(a; + a){x + a)(x + a) to n factors, or {x + a)".
Exercises
1. A building has 6 exits. In how many ways can a person leave
the building and enter by a different door?
2. A car has five seats. In how many different ways may three
people be seated, each occupying a different seat?
3. In how many different ways may 3 presents be given away to
10 people?
106. Definitions. Every distinct order in which objects
may be placed in a line or row is called a permutation, or an
arrangement. Every distinct selection of objects that can be
made, irrespective of the order in which they are placed, is called
a combination, or group.
Thus, if we take the letters a, b, e, two at a time, there are six
arrangements, namely, ab, ac, ba, be, ca, cb, but there are only
three groups, namely, ab, ac, be.
If we take the three letters all ,at a time, there are six arrange-
ments possible, namely, abc, acb, boa, baa, cab, eba, but there is
only one group, namely, abc.
Permutations and combinations are both results of mode of
selection. Permutations are selections made with the understand-
ing that two selections are considered as different even though
they differ in arrangement only; combinations are selections made
with the understanding that two selections are not considered as
different, if they differ in arrangement only.
200 ELEMENTARY MATHEMATICAL ANALYSIS [§107
In the following work, products of the natural numbers like
1X2X3; 1X2X3X4X5; etc.
are of frequent occurrence. These products are abbreviated by
the sjonbols 3\, 5 Land read "factorial three," "factorial five"
respectively.
107. Formula for the Number of Permutations of n Different
Things Taken All at a Time. We are required to find how many
possible ways there are of arranging n different things in a line.
Lay out a row of n blank spaces, so that each may receive one of
these objects, thus:
I 1 I I 2 I I 3 I I 4 I I 5 I . . . MlJ
In the fijst space we may place any one of the n objects; therefore,
that space may be occupied in n different ways. The second
space, after one object has been placed in the first space, may be
occupied in (n — 1) different ways; hence, by the fundamental
principle, the two spaces may be occupied in n(n — 1) different
ways. In like manner, the third space may be occupied in (n — 2)
different ways, and, by the same principle, the first three spaces
may be occupied in n(n — l)(n — 2) different ways, and so on.
The next to the last space can be occupied in but two different
ways, since there are but two objects left, and the last space
can be occupied in but one way by placing therein the last re-
maining object. Hence, the total number of different ways of
occupying the n spaces in the row with the n objects is the product
n(n - l)(n -2) . . 3-2-1,
or
n!.
If we use the symbol Pn to stand for the number of permutations
of n things taken all at a time, then we write
P„ = n! (1)
108. Formula for the Nxmiber of Permutations of n Things
Taken r at a Time. We are required to find how many possible
ways there are of arranging a row consisting of r different things,
§108] PERMUTATIONS AND COMBINATIONS 201
when we may 8ele(}t the r things from a larger group of n different
things.
For convenience in reasoning, lay out a row of r blank spaces,
so that each of the spaces may receive one of the objects, thus:
\ 1 I I 2 \ 3 j . . . i r-1 I I r \
In the first space of the row, we may place any one of the n objects;
therefore, that space may be occupied in n different ways. The
second space, after one object has been placed in the first space,
may be occupied in (w — 1) different ways; hence, by the fun-
damental principle, the two spaces may be occupied in n{n — 1)
different ways. In like manner, the third space may be occupied
in (n — 2) different ways; hence, the first three may be occupied
in n{n — !)(«■ — 2) different ways, and so on. The last, or rth,
space can be occupied in as many different ways as there are
objects left. When an object is about to be selected for the rth
space, there have been used (r — 1) objects (one for each of the
(r — 1) spaces already occupied). Since there were n objects to
begin with, the number of objects left is n — (r — 1), orn — r + 1,
which is the number of different ways in which the last space
in the row may be occupied. Hence, the formula:
P„,. = n(n - i)(n - 2) (n - r + i), (1)
in which P„,r stands for the number of permutations of n things
taken r at a time.
This formula, by multiplication and division by (n — r) !
becomes :
_ n(n - 1) . . . (w - r + l)(n - r){n - r - 1) . . . 3-2-1
""■ ~ {n-r){n-r-l). 3-21
n'
or P.,. = , v.- (2)
' (n — r) ! ^ '
This formula is more compact than the form (I) above, but the
fraction is not in its lowest terms.
Formula (1) is easily remembered by the fact that there are
just r factors, beginning with n and decreasing by one. Thus we
have
Pio,7 = 10X9X8X7X6X5X4.
202 ELEMENTARY MATHEMATICAL ANALYSIS [§109
Exercises ,
1. How many permutations can be made of six things taken all at a
time?
2. How many different numbers can be made with the five digits
1, 2, 3, 4, 5, using each digit once and only once to form each number?
3. The number of permutations of four things taken all at a time
bears what ratio to the number of permutations of seven things taken
all at a time?
4. How many arrangements can be made of eight things taken three
at a time?
5. How many arrangements can be made of eight things taken five
at a time?
6. How many four-figure numbers can be formed with the nine
digits 1, 2, 9 without repeating any digit in any number?
7. How many different signals can be made with seven different
flags, by hoisting them one above another five at a time?
8. How many different signals can be made with seven different
flags, by hoisting them one above another any number at a time?
9. How many different arrangements can be made of nine ball
players, supposing only two of them can catch and one pitch?
10. How many different ways may the letters of the word algebra
be written, using all of the letters?
109. Formula for the number of combinations, or groups,
of n different things taken r at a time.
It is obvious that the number of combinations, or groups, con-
sisting of r objects each that can be selected from n objects, is
less than the number of permutations of the same objects taken
r at a time, for each combination or group when selected can be
arranged in a large number of ways. In fact, since there are r
objects in the group, each group can be arranged in exactly r\
different ways. Hence, for each group of r objects, selected from
n objects, there exists r! permutations of r objects each. There-
fore, the number of permutations of n things taken r at a time, is
r! times the number of combinations of n objects taken r at a
time. Calling the unknown number of combinations x, we have
xXrl = P„„ =, ^" ,,,
{n — r)\
or solving for x
^ n\
r!(n — r)!
§109] PERMUTATIONS AND COMBINATIONS 203
This is the number of combinations of n objects taken r at a time,
and may be symbdiized
C 5J (I)
This fraction will always reduce to a whole number. It may be
written in the useful form
P _ n{n - l){n - 2) . . . (n - r + 1) ,„.
^"" ~ 1X2X3. r ' ^'''
It is easily remembered in this form, for it has r factors in both
the numerator and the denominator. Thus for the number of
combinations of ten things taken four at a time we have four
factors in the numerator and denominator, or
„ ^ 10 X 9 X 8 X 7
^">'' 1X2X3X4 ■
Exercises
1. Howmany different products of three each can be made with the
five numbers a, 6, c, d, e, provided each combination of three factors
gives a different product.
2. How many products can be made from nine different numbers,
by taking six numbers to form each product?
3. How many products can be made from nine different numbers,
by taking four numbers to form each product?
4. How many different hands of thirteen cards each can be held at a
game of whist?
6. A building has 5 entrances. In how many ways can a. person
enter the building and leave by a different door?
6. In how many ways can a child be named, supposing that there
are 400 different Christian names, without giving it more than three
names?
7. In how many ways can a committee of three be appointed from
six Italians, four Frenchmen, and seven Americans provided each
nationality is represented?
8. There are five straight lines in a plane, no two of which are par-
allel; how many intersections are there?
9. There are five points in a plane, no three of which are coUinear;
how many lines result from joining each point to every other point?
10. In a plane there are n straight lines, no two of which are parallel ;
how many intersections are there?
204 ELEMENTARY MATHEMATICAL ANALYSIS [|110
11. In a plane there are n points, no three of which are collinear;
how many straight lines do they determine?
12. In a plane there are n. points, no three of which are collinear,
except r, which are all in the same straight line; find the number of
straight lines which result from joining them.
13. In how many ways can seven people sit at a round table?
14. In how many ways can seven beads of different colors be strung
so as to form a bracelet?
15. How many different sums of money can be formed from a dime,
a quarter, a half dollar, a dollar, a quarter eagle, a half eagle, and an
eagle?
110.* The Arithmetical Triangle. In deriving by actual mul-
tiplication, as below, any power of a binomial x + a from the
preceding power, it is easy to see that any coeflSicient in the new
power is the sum of the coefficient of the corresponding term in the
multiplicand and the coefficient preceding it in the multiplicand.
Thus
x' + 3ax^ + So^a; + a'
X + a
X* + 3ax^ + 3aV + a^x
ax' + 3aV + 3a'x + a*
x'^ + Aax' + &aV + Aa'x + a\
or, retaining coefficients only, we have
1+3+3+1
1^ 1
1+3+3+1
1+3+3+1
1+4+6+4+1
from which the law of formation of the coefficients 1, 4, 6, . . .
is evideAt. Hence, writing down the coefficients of the powers
of a; + o in order, we have
§in]
PERMUTATIONS AND COMBINATIONS
205
Powers
CoefScients
]
L 2
3
4
5
6
7
8
9
10
11
0 ]
1 ]
I 1
. 2 ]
L 2
1
3 ]
I 3
3
1
4 ]
L 4
6
4
1
5 ]
L 5
10
10
5
1
6
L 6
15
20
15
6
1
7 ]
I 7
21
35
35
21
7
1
8
t 8
28
56
70
56
28
8
1
9 ]
I 9
36
84
126
126
84
36
9
1
10
L 10
45
120
210
252
210
120
45
10
1
In this triangle, each number is the sum of the number above it
and the number to the left of the latter. Thus 84 in the 9th line
equals 56 + 28, etc. The triangle of numbers was used previous
to the time of Isaac Newton for finding the coefficients of any de-
sired power of a binomial. At that time it was not suspected
that the coefficients of any power could be made without first
obtaining the coeflBcients of the preceding power. Isaac Newton,
while an undergraduate at Cambridge, showed that the coefficients
of any power could be found without knowing the coefficients of
the preceding power; in fact, he showed that the coefficients of
any power n of a binomial were functions of the exponent n.
The above triangle of numbers is known as the arithmetical
triangle or as Pascal's triangle.
111. Binomial Expansion. The demonstration of the binominal
theorem may be based upon the following law of multiplication:
The product of any number of •polynomials is the aggregate of all
the possible partial products which can be made by taking one term
and only one from each of the polynomials. This statement is
merely a definition of what is meant by the product of two or more
polynomials. (See Chapter XV, §305.) Thus,
{x + a){y + b){z + c) =
xyz ■+ ayz + bxz + cxy + abz + box + cay + abc
206 ELEMENTARY MATHEMATICAL ANALYSIS [§111
Each of vthe eight partial products contains a letter from each
parenthesis, and never two from the same parenthesis. The
number of terms is the number of different ways in which a letter
can be selected from each of the three parentheses. In the present
case this is, by §105, 2X2X2 = 8.
Let it be required to write out the value of (x + a)", where x
and o stand for any two numbers and n is a positive integer.
That is, we must consider the product of the n parentheses
(x + a)(x + a){x + a) (x + a),
by the distributive law stated above.
First. Take an x from each of the parentheses to form one of
the partial products. This gives the term x" of the product.
Second. Take an a from the first parenthesis with an x from
each of the other (n — 1) parentheses. This gives aa;""' as
another partial product. But if we take a from the second paren-
thesis and an x from each of the other (n — 1) parentheses, we get
ax"-'- as another partial product. Likewise by taking a from any
of the parentheses and an x from each of the other (n — 1) paren-
theses, we shall obtain aa;»~' as a partial product. Hence, the
final product contains n terms like ax"~', or, adding these, we
obtain nax""^ as a part of the product.
Third. We may obtain a partial product like a^x^~'^ by taking
an a from any two of the parentheses, together with the x's from
each of the other (n —2) parentheses. Hence, there are as many
partial products like o^a;»"^ as there are ways of selecting two a's,
from n parentheses; that is, as many ways as there are groups, or
combinations, of n things taken two at a time, or
n{n — 1)
r2
Hence, — - a^a;""^ is another part of the product.
1 '^
Fourth. We may obtain a partial product like a'a;"~' by taking
an a from any three of the parentheses together with the a;'s from
each of the other (» — 3) parentheses. Hence, there are as many
partial products like a'a;»-' as there are ways of selecting three o's
from n parentheses, that is, as many ways as there are combina-
§111] PERMUTATIONS AND COMBINATIONS 207
tions of n things taken three at a time, or V9^
Hence, TY^ a^x"-' is another part of the product.
In general, we may obtain a partial product like a'x"'' (where r
is an integer < n) by taking an a from any r parentheses together
with the x's from each of the other (w — r) parentheses.
Hence, there are as many partial products Uke a'x"~' as there are
ways of selecting r a's from n parentheses; that is, as many ways
as there are combinations of n .things taken r at a time, or
-r-T — '■ — Ti' Hence, -7-7 — '- — rr a'x"'' stands for any term
r] {n — r)l ' r\ (n ^ r)l •'
in general in the product (x + o)".
Finally, we may obtain one partial product like a" by taking an
a from each of the parentheses. Hence, a" is the last term in the
product.
Thus we have shown that
/ I \ 1 11 n(n — I ) , , ,
(x + a)" = X" + nax"-! -\ — ^ a^x""^ + . . .
1-2
+ r!(n°-r)l^''^""^+ ' +"^"- ^^^
This is the binomial formula of Isaac Newton. The right-hand
side is called the expansion or development of the power of the
binomial.
It is obvious that the expansion of (x — a)" will differ from the
above only in the signs of the alternate terms containing the odd
powers of a, which, of course, will have the negative sign.
Exercises
1.. Expand {u + Sy)^. Here x = u and a = 3y. By the formula
we get
u^ + bu^iSy) + I0u\3y)' + lOu^iSyy + 5u{Zyy + {3y)K
Performing the indicated operations, we obtain
u^ + 15u'y + 90u'y^ + 270u^y^ + i05uy* + 2i3yK
Expand each of the following by the binomial formula :
2.
(r«-
■ 2y.
3.
(3b -
-iy.
4.
(c +
xy.
6.
(2a;!!
-x)\
6.
(1-
ay.
7.
(-X
+ 2ay.
14.
(x^
+ x^y.
16.
(o-»-
- lyiy.
16.
(\/^ - -yaby.
208 ELEMENTARY MATHEMATICAL ANALYSIS [§112
8. (i + xy.
9. (62 - c^y.
10. (3o + iy.
11. (5d - 3yy.
12. (3a;»_- 1)*.
13. (Vo + .!;)«.
17. (a + [X + 2/1)'.
18. (a + 6 - ?/)'•
19. (a;2 + 2oa; + a^y.
112. Binomial Expansion .for Fractional and Negative
Exponents. It is proved in the Calculus that
/•. , \ 1 , , n{n — 1) , , n(m — l)(n — 2) , ,
(1 ± a;)» = 1 ± na; + -^^-^j — -x'' ± — ~ a;' + . . .
is true for fractional and for negative values of n, provided x is
less than 1 in absolute value. The number of terms in the expan-
sion is not finite, but is unlimited.
By the above formula, we have
V 1 -\- X = i. + (2) X -\ 21 ^ + " 3] x' + . . .
= 1 + (i) X - (ij X^ + (Vff) X' - (tI^)^
1
2
If --1
this becomes
V f = 1 + i - ^T + \\rs ~ "JT^" + • ■
Therefore, using five terms of the expression
\A|= 2048 ~ 1.2241 approximately.
The square root, correct to five figures, is really 1.2247. Thus the
error in this case is less than one-tenth of 1 percent if only five terms
of the series be used. The degree of accuracy in each case is depend-
ent both upon the value of n and upon the value of x. Obviously, for
a given value of n, the series converges for small values of x more
rapidly than for larger values.
As another example, suppose it is required to expand (1 — x)~'.
By the binomial theorem
(1 - x)-i = 1 + (- 1)( - x) + ~ ^ ^~,^ ~ ^\ - xy
+ -'^-'-l^^-'-'h-^y+...
= l+x+x'+x^+. . .
§113] PERMUTATIONS AND COMBINATIONS 209
If five terms of the series be used, the error is -^ f or a; «= i, or about
6 percent.
113. Approximation Fonnulas. If x be very small, the expan-
sion of
(1 + a;)« = 1 + ns + -^-^1 — x^ + ■ ■
is approximately
(1 + a;)" - 1 + nx, (1)
since x^, x' and all higher powers of x are much smaller than x.
Thus, using the symbol ^ to express "approximately equals," we
have, for example
(1.01)3 = 1.03.
For, (1 + 1/100)5 _ 1 +3/100.
The true value of (1.01)' is 1.030301, so that the approximation is
very good.
Likewise
(i - x)" ^ I — nx, (2)
if X be small.
If X, y, and z be small compared with unity, the following ap-
proximation formulas hold :
(i-+x)(i+y)^ i+x-l-y, (3)
f^-i+x-y, (4)
(i-|-x)(i-hy)(i + z)=T= i4-x-f-y-hz. (5)
The approximation formulas are proved as follows :
(1 -|- x) (1 +y) = l+x + y + xy^l+x + y, for a;?/ is small
compared to x and y.
,. I V = 1 + X — y + , , = 1 + a; — y, for the fraction is
small compared to x and y.
1 + x) (1 + y) {1 + z) ^ {1 +x + y) (1 + z) ^ 1 + X + y + z
Exercises
1. Explain the following approximation formulas, in which |x| < 1
14
210 ELEMENTARY MATHEMATICAL ANALYSIS [§113
Vl - X "5=
(1 +x)-i^
(1 + x)-i =F
(1 +x^)i ===
2. Compute the approximate numerical value of the following :
(a) (1.03) i (d) (1.05) i
(6) (1.02) (1.03) (e) 1.02/1.03
(c) (1.01)(1.02)/(1.03)(1.04).
3. The formula for the period of a simple pendulum is
T =WT7i-
For the value of gravity at New York, this reduces to
T =
6.253'
in which I, the length of the pendulum, is measured in inches. This
pendulum beats seconds when
I = (6.253)=i or 39.10 inches.
What is the period of the pendulum if I be lengthened to 39.13 inches?
Hint:
T =
6.253
^ - "6:253" - 6:253^^ + ^^^
VT
(1 + h/2l).
6.253
Take I = 39.10, and h = 0.03.
Then
?" = 1 +■ 0.03/78.20
= 1.00038.
A day contains 86,400 seconds. The change of length would, there-
fore, cause a loss of 32.8 seconds per day, if the pendulum were
attfiched to a clock,
§114] PERMUTATIONS AND COMBINATIONS 211
4, On the ocean how far can one see at an elevation of h feet above
its surface?
Call the radius of the earth o( = 3960 miles), and the distance one
can see d, which is along a tangent from the point of observation to
the sphere. Since h is in feet, and since a + Toon! d, and a are the
sides of a right triangle, we have (o + ^/5280)'' = d' + a\
or
"[
' + sis] ■-''■+«■■
Expanding the binomial by the approximation formula we have
.[
'+mk] ='' + <
d2 = 2a;i/5280
= 2 X 3960^/5280
-¥,
or
d = Vp
where d is expressed in miles and h in feet. See §68, exercise 13.
5. By what percent is the area of a circle altered if its radius of
100 cm. be changed to 101 cm.?
6. By what percent is the volume of a sphere, |-7ro', altered if the
radius be changed from 100 cm. to 101 cm.?
7. If the formula for the horse power of a ship is I.H.P. = „„-,i
where S is speed in knots and D is displacements in tons, what in-
crease in horse power is required in order to increase the speed from
fifteen to sixteen knots, the tonnage remaining constant at 5000?
What increase in horse power is required to maintain the same speed
if the load or tonnage be increased from 5000 to 5500?
114.* Graphical Representation of the Coefficients of any
Power of a Binomial. If we erect ordiaates at equal intervals
on the X-axis proportional to the coeflBcients of any power of a
binomial, we find that a curve is approximated, which becomes
very striking as the exponent is taken larger and larger. In Fig.
97 the ordinates are proportional to the coefficients of the 999th
power of {x + a). The drawing is due to Quetelet.
212 ELEMENTARY MATHEMATICAL ANALYSIS [§114
The limit of the broken line at the top of the ordinates in Eig.
97 is, as n is increased indefinitely, a beU-shaped curve, known as
Fig. 97. — Graphical representation of the values of the binomial
coefficients in the 999th power of a binomial. The middle coeflScients
are taken equal to 5, for convenience, and the others are expressed
to that scale also.
the probability curve. In treatises on the Theory of Probability,
it is shown that the equation of the curve is 2/=ae~*^^
CHAPTER VIII
PROGRESSIONS
116. An Arithmetical Progression or an Arithmetical Series,
is any succession of terms such that each term differs from that
immediately preceding by a fixed number called the common
difference. The following are arithmetical progressions:
(1) 1, 2, 3, 4, 5.
(2) 4, 6, 8, 10, 12.
(3) 32, 27, 22, 17, 12.
(4) 2i, 3i 5, 6i 7i.
(5) (u - v), u, {u + v).
(6) a, a + d, a + 2d, a -\- 3d, . . .
The first and last terms are called the extremes, and the other
terms are called the means.
Where there are but three numbers in the series, the middle
number is called the arithmetical mean of the other two. To
find the arithmetical mean of the two numbers a and 5, proceed as
follows:
Let A stand for the required mean; then, by definition
A — a = b — A,
whence
A - ^ + ^
Thus, the arithmetical mean 6f 12 and 18 is 15, for 12, 15, 18 is an
arithmetical progression of common difference 3.
By the arithmetical mean, or arithmetical average, of several
numbers is meant the result of dividing the sum of the numbers
by the number of the numbers. It is, therefore, such a number
213
214 ELEMENTARY MATHEMATICAL ANALYSIS [§116
that if all numbers of the set were equal to the arithmetical mean,
the sum of the set would be the same.
The general arithmetical progression of n terms is expressed by:
Number of
term: 12 3 4 . n
Progression: a, (a + d), (a + 2d), (a + 3d), . . . (a + [n — 1] d)
Here a and d may be any algebraic numbers whatsoever, integral
or fractional, rational or irrational, positive or negative, but n
must be a positive integer. When the common difference is nega-
tive, the progression is said to be a decreasing progression ; other-
wise, it is an increasing progression.
From the general progression written above, we see that a for-
mula for the nth term of any arithmetical progression may be
written
I = a -H (n - i)d, (1)
in which I stands for the nth term.
Formula (1) enables us to obtain the value of any one of the num-
bers, I, a, n, d, when the other three are given. Thus:
(1) Find the 100th term of
3 4- 8 -h 13 -I- . . .
Here a = 3, d = 5, n = 100.
Therefore ' Z = 3 + 99 X 5 = 498.
(2) Find the number of terms in the progression
5 + 7 -I- 9 + . . . + 39.
Here a = 5, d = 2,1 = 39.
Therefore 39 = 5 -t- (ra - 1)2,
or n = 18.
(3) Find the common difference in a progression of fifteen terms in
which the extremes are f and 425.
Here u, = ^,1 = 42^, n = 15,
whence 42| = J -F (15 - l)d,
or d = 3.
116. The Sum of n Terms. If s stands for the sum of n terms
of an arithmetical progression, and if the sum of the terms be
§116] PROGRESSIONS 215
written first in natural order, and again in reverse order, we have
s = a + (a + d) + (a + 2d) + + (a + [n - 1] d), (1)
s = 1+ {I - d) + {I - 2d) + . . + Q -In- l]d). (2)
Adding (1) and (2), term by term, noting that the positive and
negative common differences nullify one another, we obtain
2s = (a + Z) + (a + Z) + (a + 0 + . ■ ■ + (a + l), (3)
or, since the number of terms in the original i5rogression is n, we
may write
2s = n{a + I),
or s = n(a + l)/2. (4)
If the value for I, from (1) §115, be substituted in formula
(4) it becomes
s = n [2a + (n - i)d]. (5)
In equation (4), (a + Z)/2 is the average of the first and nth
terms. The formula (4) states, therefore, that the sum equals the
number of the terms multiplied by the average of the first and last.
An arithmetical progression is a very simple particular instance
of a much more general class of expressions known in mathematics
as series. A series is any sequence of terms formed accord-
ing to some law, such as:
(x + 1) + (x + 2y+ {x + sy +. . .
x + 3x^ + 5x^+ . . .
cos X + cos 2x + cos 3x -\- . . .
It is only in a very limited number of cases that a short expression
can be found for the sum of n terms of a series. An arithmetical
progression is one of these cases.
Formula (4) enables us to find the value of any one of the numbers
s, n, a, I, when the values of the other three are given. Thus:
(1) Find the number of terms in an arithmetical progression in
which the first term is 4, the last term 22, and the sum 91.
Here a = 4, Z = 22, s = 91,
whence, 91 = ra(4 + 22) /2,
or n = 7.
216 ELEMENTARY MATHEMATICAL ANALYSIS [§116
The two formulas, (1) §116 and (4) §118, contain five letters;
' hence, if any two of them stand for unknown numbers, and the values
of the others are given, the values of the two unknown numbers can be
found by the solution of a system of two equations. Thus :
(2) Find the number of terms in a progression whose sum is 1095, if
the first term is 38 and the difference is 5.
Here ■ s = 1095, a = 38, and d = 5,
whence, I = 38 + {n - 1)5, (6)
1096 = n(38 + l)/2. (7)
From (6) / = 33 + 5n. (8)
From (7) 2190 = 38ra + nl. (9)
Substituting the value of / from (8) in (9), we get
2190 = 71n + 5nK (10)
Solving this quadratic equation, we find
n = 15, or - 29.2.
The second result is inadmissible, since the number of terms cannot
be either negative or fractional.
Exercises
Solve each of the following:
1. Given, o = 7, d = 4, n = 15; find 2 and s.
2. Given, a = 17,1 = 350, d = 9; find n and s.
3. Given, a = 3, n = 50, s = 3825; find I and d.
4. Given, s = 4784, a = 41, d = 2; find Zand n.
5. Given, s = 1008, d = 4, Z = 88; find a and n.
6. Find the sum of the first n even numbers.
7. Find the sum of the first n odd nvmibers.
8. Insert nine arithmetical means between —7/8 and + 7/8.
9. Sum (o + 6)2 + [a" + ¥) + (,a -byton terms.
10. Find the sum of the first fifty multiples of 7.
11. Find the amount of $1.00 at simple interest at 5 percent for
1920 years.
12. How long must $1.00 accumulate at 3| percent simple interest
until the total amounts to $100?
13. How many terms of the progression 9 + 13 + 17 + . .
must be taken in order that the sum may equal 624? How many
terms must be taken in order that the sum may exceed 750?
§117] PROGRESSIONS 217
14. Show that the only right triangles whose sides are in arithmetical
progression are those whose sides are proportional to 3, 4, and 6.
117. A geometrical progression or a geometrical series is any
succession of terms such that each term is the product of the
preceding term by a fixed factor called the ratio. The following
are examples:
(1) 3, 6, 12, 24, 48. (3) 1/2, 1/4, 1/8, 1/16, 1/32.
(2) 100, -50, 25, -12i (4) a, ar, ar\ ar\ ar* . .
The geometrical mean G of two numbers, a and 6, is a number-
such that a, G, 6 is a geometrical progression. By definition
G/a = b/G,
whence,
G^ = ab,
or
G = Vab.
Thus, 4 is the geometrical mean of 2 and 8. The arithmetical
mean of 2 and 8 is 5. The geometrical mean of n positive num-
bers is the value of the nth root of their product. Thus the geo-
metrical mean of 8, 9, and 24 is -?/ 8 X 9 X 24 = 12.
118. The nth Term and the Sum of n Terms. If a represents
the first term and r the ratio of any geometrical progression, the
progression may be written:
Number of term: 123 4 .. n— 1 n.
Progression: o, ar, ar^, ar', . . ar"'^, ar"~^.
Therefore, representing the nth term by I, we obtain the simple
formula
1 = ar»-i. (1)
Representing by s the sum of n terms of any geometrical pro-
gression, we have
s = a -\- ar + ar^ + . . . + ar" ~^ + ar" ~ ^,
or,
s = ail+r + r^+ . . + r"-^ + r"-^).
But, by a fundamental theorem in factoring, ^ the expression in the
1 See Appendix, Chapter XV.
218 ELEMENTARY MATHEMATICAL ANALYSIS [§118
(2)
parenthesis is the quotient of 1 — r» by 1 — r. Hence,
a(i — r»)
Another form is obtained by introducing I by the substitution
or»-' = I,
a — rl
which gives s = — — — (3)
Formula (1), or (2), enables one to find any one of the four
numbers involved in the equations when three are given. The
two formulas (1) and (2) considered as simultaneous equations
enable one to find any two of the five numbers a, r, n, I, s, when the
other three are given. But if r be one of the unknown numbers,
the equations of the system may be of a high degree and beyond
the range of Chapter VI unless solved by graphical means. If
n be an unknown number, an equation of a new type is introduced,
namely, one with the unknown number appearing as an exponent.
Equations of this type, known as exponential equations, will be
treated in the chapter on logarithms. The following examples
illustrate cases in which the resulting single and simultaneous
equations are readily solved.
(IJ Insert three geometrical means between 31 and 496.
Here
a = 3l,l = 496, and n = 6.
Hence,
496 = 31 X r'
r* = 16,
or
r = ± 2.
Consequently the required means are either 62, 124, and 248, or — 62,
+ 124, and - 248.
(2) Find the sum of a geometrical progression of five terms, the
extremes being 8 and 10,368.
Here
a = 8,1 = 10,368, and n = 5.
Hence,
10,368 = 8r* (1)
§118] PROGRESSIONS 21,9
aad
8 = (10,368r - 8)/{r - 1). (2)
From the first,
r = 6
whence, from the second,
s = 12,440.
(3) Find the extremes of a geometrical progression whose sum is 635,
if the ratio be 2 and the number of terms be 7.
Here
s = 635, r- = 2, and n = 7.
Hence,
I = a2«, (3)
635 = (2/ - a). (4)
Substituting I from (3) in (4), we get
635 = 128 o - a.
Hence,
a = 5, and I = 320.
(4) The fourth term of a geometrical progression is 4, and the
sixth term is 1. What is the tenth term?
Here
ar^ = 4, (5)
and
ar^ = 1. (6)
Dividing (6) by (5) we obtain
r
2 _ 1
i, or r = + i
Therefore, from (5),
a = 4^/r^ = +32.
Then the tenth term is
± 32(+ \y = tV.
Exercises
1. Find the sum of seven terms of 4 + 8 + 16 + . . .
2. Find the sum of - 4 + 8 - 16 + . . . to six terms.
3. Find the tenth term and the sum of ten terms of 4 — 2 + 1 ■
4. Find r and s; given a = 2,1 = 31,250, Ji = 7.
220 ELEMENTARY MATHEMATICAL ANALYSIS [§119
6. Insert two geometrical means between 47 and 1269.
6. Insert three geometrical means between 2 and 3.
7. Insert seven geometrical means between o' and 6*.
8. Show that the quotient (o" — 6»)/(o — 6) is a geometrical
progression.
9. Sum x"""^ + x"~' y + a;""' y' + . . to n terms.
10. Sum a;"~i — a^~' y + s""' y' — ■ ■ . to w terms.
11. Sum a + ar~^ + ar~' + . . . to n terms.
12. If a, b, c, d, ' . . . are in geometrical progression, show that
a'^ + 6', 6* + c^, c^ + (i^ . . are also in geometrical progression.
13. If any numbers are in geometrical progression, show that their
differences are also in geometrical progression.
14. A man agreed to pay for the shoeing of his horse as follows:
1 cent for the first naU, 2 cents for the second nail, 4 cents for the third
nail, and so on until the eight naUs in each shoe were paid for. What
did the last nail cost?. How much did he agree to pay in all?
119. Compound Interest. Just as the amount of principle and
interest of a sum of money at simple interest for n years is ex-
pressed by the (n + l)st term of an arithmetical progression, so,
in a similiar way, the amount of any sum at compound interest for
n years is represented by the (n + l)st term of a geometrical pro-
gression. Thus, the amount of $1.00 at compound interest at
4 percent for twenty years is given by the expression
1(1.04)2".
The amount of p dollars for n years at r percent is
K' + i5-o)"-
The present value of $1.00, due twenty years hence, estimating
compound interest at 4 percent, is
1/(1.04)2".
The value of $1.00, paid annually at the beginning of each year
into a fund accumulating at 4 percent compound interest, is, at
the end of twenty years
(1.04)1 + (104)'' + . . . (1.04)2",
which is the sum of the terms of a geometrical progression of
twenty terms.
§120] PROGRESSIONS 221
Problems of this character in compound interest, in compound
discount, and in the more complicated problems that proceed
therefrom, are basal to the theory of annuities, life insurance, and
depreciation of machinery and structures. The computation of
the high powers involved necessitates the postponement of such
problems until the subject of logarithms has been explained.
120. Infinite Geometrical Progressions. If the ratio of a
geometrical progression be a proper fraction, the progression is
said to be a decreasing progression. Thus,
1 1 i i 1 cnA ill 1
■I) 2) i! 8) iw> ana 3, s, jt, ^t
are decreasing progressions. If we increase the number of terms
in the first of these progressions the sums will always be less than
2; but the difference (2 — s) will become and remain less than any
pre-assigned number.
Definition: A constant, a, is called the limit of a variable,
t, if, as t runs through a sequence of numbers, the difference
(a — t) becomes, and remains, numerically smaller than any
pre-assigned number.
By definition, 2 is, therefore, the limit of the first of the above
progressions. The sum of n terms of this particular progression
should be written down by the student for a number of successive
values for n, thus:
Number of terms:
1, 2, 3, 4, 5, ... 10,
Sum: 1, 1 + i 1 + f , 1 + I, 1 + li . . 1+Ui,
The nth term differs from 2 by only l/2»-i.
It is easy to show that the sum of every decreasing geometrical
progression approaches a fixed limit as the number of terms
becomes infinite. Write the formula'^
in the form
If we suppose that r is a proper fraction and that n increases with-
s
=
a —
or»
1 -
- r
0
ar'
222 ELEMENTARY MATHEMATICAL ANALYSIS [§121
out limit, then r» can be made less than any assigned number; for,
the value of any power of a proper fraction decreases as the ex-
ponent of the power increases. As the other parts of the second
fraction in (1) do not change in value as n changes, the fraction
as a whole can be made smaller than any number that can be
assigned. Hence, we write
limit
n— 00
b]'Th <^'
The left-hand side is read: "The limit of s as n becomes
infinite." The symbol = means: "approaches" or "becomes."
Exercises
As n = 00 , find the limit of each of the following :
1. * - i + 1 - tV + • •
Here
a = -^jT = — -3,
1
whence, limit s = — j-r = f .
1 ~ ( "2)
2. 0.3333 . .
Here a = -,%, r = ,Vi
3
whence, limit s = — = \
3. 9-6+4- ' ^.\-\+-h----
4. 0.272727 ... 7. 4 4- 0.8 + 0.16 + . . .
5. 0.279279279 . . .
8. Express the number 8 as the sum of an infinite geometrical
progression whose second term is 2.
121.* Graphical Representation. Note that all the essentials
of a geometrical progression may be studied if we assume the
first term to be unity, for the number a occurs only as a single
constant multiplier in each term, and also occurs in the same
manner in the formulas for I and s.
To represent the geometrical series 1 + r + r'' . . + r"-'
graphically, lay off OM = 1 on OY, OSi = 1 on OX, SiPi =
r on the unit line, and draw MP^. Draw the arc P11S2 and erect
P2S2. Draw the arc P2'S2 and erect PzSs. Continue this con-
struction until the perpendicular P„iS„ is erected. The series of
trapezoids OMPiSi, S1P1P2S2, SJ'^iPiSi, .. . , S,_iP„_iP„S„
§121]
PROGRESSIONS
223
are similar and, since PiSi = r X OM, it follows that P2S2 =
rPiSiyPiSa = rP^Si, . , P„S„ = rF„_iS„_i. Hence we have:
OM = OSi =1
PkSi = S1S2 = r .'. 0<Si2 = 1 + r = sum of 2 terms
P2S2 = O203
,0^3 = 1 +r + r2
sum of 3 terms
PzSs = S^Si = r^ ;. OSi = 1 + r + r^ + r' = sum of 4 terms
Pn-iSn-i = Sn-iS„ = r"-' .-. OSn = 1 + r +r' +
sum of n terms.
*m— 1 =
O Si S2 S3 S« Sii
Fig. 98. — Graphical construction of the sum of a G. P. r > 1.
Fig. 98 shows the series whose ratio is r- = 1.2. Fig. 99 shows
the series whose ratio is 0.8.
Y
U
M
P,
P2
Pa
1
^ \
,.X
r^
f*
Pi
'
1
^
n
NJTs —- —
- — .___^
O Si S, Sa Si So L
Fig. 99. — Graphical construction of the sum of a G. P. r < 1.
The line MPi has the slope (r - 1) in Fig. 98 and the slope
— (1 — r) in Fig. 99. In each case the F-intercept is 1. Its
1-2/
equation is, in both cases, y = (r
In both figures, when y = P^S,
l)x + 1, ora; =
1 -r
r", X = OSn. Substituting
these values for x and y, we get for the sum of n terms,
1 — r"
Fig. 98 shows that when the number of terms is
1 -r
224 ELEMENTARY MATHEMATICAL ANALYSIS [§122
allowed to increase without limit, the, sum OSn also increases with-
out limit. Fig. 99 shows that when the number of terms is made to
increase without limit, the sum 0S„ approaches OL as a limit.
Now the value of OL is the value of x when ^ = 0. Hence the
limit of the sum of the progression, or OL, is -t-^ —
Consult also §9, problem 6, exercise 3 and Figs. 15, 16.
122.* Harmonical Progressions. A series of terms such that
their reciprocals form an arithmetical progression are said to form
an harmonical progression. The following are examples:
C1^ 1 1 1 1
('■) 2: 3> 4! T-
(2) 1, T, T) TT-
(3) l/{x-y),l/x, l/(x + tj).
(4) i 1, - 1, - i
(5) 4, 6, 12.
(6) 1/a, l/(a + d), 1/ia + 2d), . .
Although harmonical progressions are of such a simple character,
no simple expression has been found for the sum of n terms. Our
knowledge of arithmetical progressions enables us to find the
value of any required term and to insert any required number
of harmonical means between two given extremes, as in the
examples below.
(1) Write six terms of the harmonical progression 6, 3, 2.
We must write six terms of the arithmetical progression, ^, ^, ^.
The common difference of the latter is ^, so that the arithmetical pro-
gression is §, §, §, f , ^, 1, and the harmonical progression is 6, 3, 2,
1.5, 1.2, 1.
(2) Insert two harmonical means between 4 and 2.
We must insert two arithmetical means between ^ and -^; these are
^ and -1%, whence the required harmonical means are 3 and 2.4.
123.* Harmonical Mean. The harmonical mean of two
numbers is found as follows: Let the two numbers be a and 6
and let H stand for the required mean. Then we have
1/H - 1/a = 1/6 - 1/H.
That is,
2/H = 1/a + 1/6 = (a -I- 6) /ab.
Hence,
• H = 2ab/(a -1- b). (1)
§124] PROGRESSIONS 225
Thus the harmonical mean of 4 and 12 is 96/(4 + 12) = 6.
By the harmonical mean of several numbers is meant the reciprocal
of the arithmetical mean of their reciprocals. Thus the har-
monical mean of 12, 8, and 48 is 13i-t-
124. * Relation between A, G, and H. As previously found,
A= {a+ 6)/2, G= V^,H = 2ah/{a + b).
Hence,
AH = ab,
and, since ab = C,
AH = G\
or
G = VaH. (1)
Exercises
1. Continue the harmonical progression 12, 6, 4.
2. Find the difference (1.8 + 1.2 4- 0.8 + . to 8 terms)
- (1.8 + 1.2 + 0.6 + . . . to 8 terms).
3. If the arithmetical mean between two numbers be 1, show that
the harmonical mean is the square of the geometrical mean.
Questions and Exercises for Review of Chapters I to VIII
1. Define scale; uniform scale; non-uniform scale; arithmetical
scale; algebraic scale; double scale.
2. Define constant; variable.
3. Define function; increasing function; decreasing function; even
function; odd function.
4. Give illustrations of even functions; of odd functions.
6. Express the area, A, of an equilateral triangle as a function of the
length, X, of its sides.
6. Express the volume, V, of a right circular cone as a function of its
altitude h. The radius of the base is 10 inches.
7. A strip of tin L feet long and 40 inches wide is made into a gutter
with rectangular cross section, by bending up an equal portion of each
side. Express the cross section, y, of the gutter as a function of the
breadth, x, of the amount of tin turned up. Show that the maximum
cross section is 200 square inches.
8. A strip of tin 24 inches square has an equal square cut from each
corner. The rectangular projections are then turned up to form a tray
15
226 ELEMENTARY MATHEMATICAL ANALYSIS [§124
with square base and rectangular sides. If x is the side of the square
cut out show that 4x(12 — x)* is the function representing the volume
of the tray.
9. In a triangle whose sides are 6, 8, and 10 feet is inscribed a rec-
tangle the base of which lies in the longest side of the triangle. Ex-
press the area, A, of the rectangle as a function of its altitude, h.
10. A ladder 20 feet long leans against the vertical wall of a house.
Express the area, A, of the triangle formed by the ladder, the wall, and
the horizontal ground, as a function of the distance, x, of the foot of
the ladder from the wall.
11. Find graphically the values of the following: (a) (31.6) (7.21);
(6) f^; (c) (1.36)'; (d) ~-y
12. Describe the method of representing the position of points on a
plane by the rectangular, or Cartesian, system of coordinates. Define
axes; origin; abscissa; ordinate; quadrant. How are the quadrants
numbered?
13. What is meant by the graph, locus, or curve, of an equation?
14. What is meant by the equation of a curve, graph, or locus of a
point.
15. Which of the following points are on the curve Zx -\-2y = 4:
(a) (2, -1); (6) (3, 1); (c) (-4, 8); (d) (0, 0).
16. Find the distance of each of the following points from the origin :
(a) (1, 3); (6) (-2, 3); (c) (2, -3); (d) (-3, -2).
17. Show that, for all values oi m, y = mx is a straight line passing
through the origin.
18. Show that the equation of any straight line passing through the
origin is of the form y = mx.
19. Find the equation of a straight line passing through the origin
and the point (—3, 5).
20. Show that, for all values of m and b,y = mx + 6 is the equation
of a straight line.
21. Show that the equation of any straight line is of the form
y = mx + b.
22. Find the equation of a straight line passing through the points
(1, 3) and (-2, 5).
23. Define slope of a straight line.
24. Define K-intercept, and X-intercept, of a straight line.
25. Find the slope, s-intercept, and 2/-intercept, for the following:
(a) 3x +2y = 6; (6) x - 2y = 5; (c) 2y - 3x = 7.
26. Define X-, and K-intercepts of a curve.
§124] PROGRESSIONS 227
27. Write the equations of a line if:
(a) F-intercept is 3 and slope is 2,
(6) y-intereept is 1 and slope is —2,
(c) y-intercept is —2, and slope is 5,
id) X-interoept is 3 and slope is 2,
(e) X-intercept is —2 and slope is 3,
(J) X-intercept is J and slope is — ^,
(g) X-intercept is 2 and i/-intercept is 3.
28. What is meant by curve of the parabolic type?
29. What is meant by curves of the hyperbolic type?
30. What is the parabola?
31. What is the equilateral, or rectangular, hyperbola?
32. What is the cubical parabola?
33. What is the semi-cubical parabola?
34. When is a curve symmetrical with respect to the X-axis; with
respect to the y-axis; with respect to the line x = y; with respect to
the line y = — x; with respect to the origin? Give equation of two
curves for each of the cases considered above.
35. Sketch, y = x^; y = \x^; y = 2x\
36. Sketch y'' = x; y^ = Jx; y' = 2x.
37. Sketch y = x'; y = - x'.
1 2 1
38. Sketch y = -; V = - ^: y = 2i'
39. Sketch x'^ = y'; x' = y'^.
40. Sketch y = x^; y = — x'.
41. Define rational equation, empirical equation.
42. Write the equation of the curve y = x^ — 3x, after it is
translated
(a) two units to the right; (6) three units to the left; (c) one unit
up; (d) five units down; (e) one unit to the left and two units down.
43. Find the coordinates of the vertex of :
(o) y = x'' + 2x; (6) y = x' - 2x + 3;
(c) y = 3x^ + 6x; (d) y = 6x - 3x^ + 2.
2; _]_ 3
44. Show that y = — 3^ is an hyperbola. Write the equation of
its asymptotes.
45. What is meant ty shearing notion?
46. Show that shearing the curve y = ax' in the line y = mx, is
equivalent to translating the original curve. Find the coordinates
of the vertex of the translated curve.
47. What is meant by the roots of a function?
228 ELEMENTARY MATHEMATICAL ANALYSIS [§124
48. Find the roots of:
(a) *2 + 2x - 3; (6) x^ - 3x; (c) 3x^ + 2x - 6.
49. Write the equation of the curve j/^ = x' — x' when reflected in :
(a) the X-axis; (6) the F-axis; (c) the line x = y; (d) the line
X = — y.
50. The roots of a function correspond to what points on the graph
of the function?
52. Write the equation of a circle, radius o, center at the origin;
center at the point (h, k).
53. Show that x' + y' + 2gx + 2/v + d = 0 is a circle.
54. Find the coordinates of the center and the length of the radius
of:
(a) x' + y' - 2x - ^y + 1 = 0; {d) 2x' + 2y^ + 3x + by = 0;
(6) x2 + ^2 + 2x + 42/ + 1 =0; (e) Sx^ + S?/^ - 6x - V2y = 10;
(c) x2 + 2/2 + 3x - 42/ = 0; (/) x^ + 2/' + 7x - VZy = 25.
66. Which circles of exercise 54 pass through the origin?
66. Write the equation of the circle if i
(o) the radius is 5 and the center is at (1, 2) ; '
(6) the radius is 6 and the center is at ( — J^, 2) ;
(c) the radius is 10 and the center is at ( — 2, — 3) ;
(d) it passes through the origin and the center is at (1, 1) ;
(e) it passes through the origin and the center is at ( — 2, 3);
(/) it passes through (1, 2) and the center is at (—2, 3).
57. Write the equation of a line passing through the origin and the
center of the circle '
x2 + 2/2 - 2x + 32/ = 5.
58. Write the equation of a circle passing through the point (2, 3)
and through the center of the circle
x2 + 2/^ - 3x - 22/ = 0.
69. Show that if two straight lines are mutually perpendicular, the
slope of one is the negative reciprocal of the slope of the other.
60. Show that {x - a)^ + y^ = a^ and (x - ZaY + y^ = a' are
tangent to each other.
61. Find analytically the coordinates of the points of intersection
of x2 + 2/2 — 4x — 92/ = 9 and y — ^ x + 1.
62. Find approximate solutions for exercise 61 by drawing the
curves on squared paper.
63. Solve graphically the simultaneous equations
x^ + y' - 2x ~ 4:y = 4:
-£2 + 2/2 + 4x - 42/ = 0.
64. Define degree'; radian.
66. Define the six circular functions.
§124] PROGRESSIONS 229
66. Express the following as radians:
(o) 45°; (6) 90°; (c) 180°; (d) 135°; (e) 225°; (f) 60°; (g) 30°; (h) 300°;
(i) 270°; U) 315°; (ft) 120°; (l) 160°; (ra) 216°; (n) 310°.
67. Express the following radians as degrees:
(a) i^; (fe)i^; (c) |^; (rf) |^; (e) |^; (/) 3; (?) 2.
68. How many revolutions per minute are 10 radians per second?
5 7r radians per second? fir radians per second?
69. A car is running at the rate of 30 miles per hour. Its 36-inch
tire is revolving at the rate of how many radians per second?
70. A shaft rotates at the rate of 15,000 revolutions per minute.
What is its angular velocity in radians per second?
71. Give the values of the circular functions of:
(a) 30°; (6) 60°; (c) 45°.
72. Give the algebraic signs of the functions of an angle in the first
quadrant; in the second quadrant ; in the third quadrant; in the fourth
quadrant.
73. Give the functions of the following angles:
(o) 120°; (6) 135°; (c) 150°; (d) 210°; (e) 225°; (J) 240°; {g) 300°;
(h) 316°; W 330°; (j) 0°; (k) 90°; (I) 180°; (m) 270°; (n) 360°.
74. Find the functions of a if :
(a) sin a = f and cos a is negative;
(6) sin a = f and cos a is positive;
(c) sin a = -f and tan a is positive;
(d) sin a = f and tan a is negative;
(e) tan a = 2 and cos a is negative;
(/) tan a = — 3 and sin a is positive;
(g) sec or = 5 and tan a is negative.
75. Which of the circular functions are even functions? Which are
odd functions?
76. Show that cos a = sin (^ — a) .
77. Draw the graph oi y = sin x.
78. Show that the curve for y = cos x may be obtained from the
curve for y = sinx by translating it ■ir/2 units to the left.
79. Show that sin^ a + cos^ a = 1.
80. Show that sec^ a = 1 + tan" a.
81. Show that esc" a = 1 + cot" a.
82. Show that tan a =
cos a
nn on ii i J cos a
83. Show that cot a = -^ •
sm a
230 ELEMENTARY MATHEMATICAL ANALYSIS [§124
84. Express sin a; in terms of:
(a) cos x; (6) tan x; (c) cot x; (d) sec x; (e) esc x.
86. Express cos x in terms of:
(a) sin x; (6) tan x; (c) cot x; {d) sec x; (e) esc x.
86. Express tan x in terms of:
(a) sin x; (6) cos x; (c) cot a:; (d) sec x; (?) esc a;.
87. The longer leg of a plot of land in the form of a 60° right
triangle is 80 rods. Find the area of the plot in acres.
88. A plot of land in the form of a 60° right triangle contains
72 acres.
Find the length in rods of each side of the triangle.
Hint : Let x represent the number of rods in the length of the shorter
leg.
89. The shorter side of a rectangle is 100 feet, the diagonal is 200
feet. Find the length of the longer side.
90. Explain how points may be located in a plane by means of polar
coordinates.
91. Define pole; polar axis; radius vector; vectorial angle.
92. Draw curves for:
(a) p = 1; (6) p = 2; (c) p = 3; (d) p = 5; (e) e = 0; (/) e =,r/4;
(g) e = 7r/3; {h) e = T-/2; 6 = 2.
Hint: fl is measured in radians.
93. What curve is represented by p = a cos 9? Prove.
94. What curve is represented by p = 6 sin 9? Prove.
95. Draw on a sheet of polar coordinate paper curve for the
following :
(a) p = 2 cos e; (6) p = — 2 cos 9;
(c) p = 2 sin 9; (d) p = — 2 sin 9.
96. Prove that p = a cos 9 + 6 sin 9 is a circle.
97. Draw curves for the following :
(a) p = 2 cos 9 + 3 sin 9; (6) p = 3 cos 9 — 2 sin 9;
(c) p = — 2 cos 9 + sin 9; (d) p = — 3 cos 9 — 3 sin 9.
98. Draw the circles p = 1 and p = cos 9 and from them plot the
graph for p = 1 + cos 9.
99. Plot curve for the following equations :
(a) p = 1 + sin 9; (6) p = 1 — sin 9;
(c) p = 2 + cos 9; (d) p = 1 - 2 cos 9.
§124] PROGRESSIONS 231
100. From a sheet of polar coordinate paper, form M3, find values
for the following :
(a) sin 30°; (6) cos 30°; (c) sin 45^; (d) cos 45°; (e) tan 45°; (/) cos 10°;
(g) sin 116°; (h) cos 216°; (i) sin 127°; (j) tan 37°; (fc) sin 227°;
{I) cos 316°.
101. Show that when the curve for p = f{8) is rotated about the
pole through an angle a, its equation becomes p = f{e — a).
102. State fourteen "Theorems on Loci."
103. Find the polar equation of a straight line.
104. The center of the circle p = 10 sin (9 — a) lies on the line
X — y = 3. Find a.
105. The center of the circle p = 10 cos (0 — a) lies on the line
3x - 2y = 1. Find a.
106. The center of the circle p = 5 sin {$ + a) lies on the line
X - 2?/ = 6. Find a.
107. Write the Cartesian equations for:
(a) p = 2 cos e + 3 sin 6;
(6) p = 3 cos 9 — 2 sin $;
(c) p = 2 sin 9 — 3 cos $.
108. Solve analytically 2 = 2 cos 9 — 3 sin 9 for all values of 8
between 0° and 360°.
109. Solve graphically the equation given in exercise 108.
110. Sketch a curve for y = - — 2x.
" X
111. Sketch a curve for y = ^ -j- sin x.
112. A circle is inscribed in a 30°, 60° right triangle. Find the
diameter of the circle if the shorter leg of the triangle is 4 inches; if
the longer leg of the triangle is 6 inches; if the hypotenuse of the
triangle is 10 inches. Find the lengths of the three sides of the tri-
angle if the radius of the inscribed circle is 6 inches.
113. A circle is inscribed in a 45° right triangle. Find the diameter
of the circle if the legs of the triangle are each 4 inches in length.
114. A circle is circumscribed about a 30°, 60° right triangle. Find
the radius of the circle if the hypotenuse of the triangle is 10 inches.
116. Write the polar equation for
x^ - y^ = a2(x2 + 2/2)2.
116. Define an ellipse; major axis; minor axis.
117. Give parametric equations for an ellipse.
232 ELEMENTARY MATHEMATICAL ANALYSIS [§124
118. Find the coordinates of the center and the lengths of the semi-
axes of the ellipse
X = 3 + 2 cos a
y = 2 — svD. a.
119. Show that every section of a right circular cylinder by a plane
is an ellipse.
120. Show that the projection of a circle upon a plane is an ellipse.
121. Define an equilateral, or rectangular,\ hyperbola. Define an
hyperbola.
122. Give parametric equations for an hyperbola.
123. Define the axes, the center, and the asymptotes of an
hyperbola.
124. Find the coordinates of the center, the lengths of the semi-
axes, and the equations of the asymptotes for
x' - y^ + 2x - ^y = 11.
126. Find the equation for the curve of sy = 4 when rotated about
the origin through an angle of —45°
126. Define conjugate hyperbolas.
127. Write the equation of the hyperbola conjugate to
x'i — y'' — X -\- iy = 11.
128. Sketch the curve with asymptotes for
X = 3 + 2 sec a
2/ = 1 — 3 tan a.
129. Write the equation of the curve formed when the circle
x2 -|- j/2 = o' is sheared in the Une y = x. Sketch the curve.
130. Write the equation of the curve formed when the hyperbola
x^ V^
-J — ^ = 1 is sheared in the line y = x. Sketch the curve.
131. State and prove the remainder theorem.
132. State and prove the factor theorem.
133. Without performing the division, find the rehiainder of
(s' - 2x2 + 3 - 1) -H (a; + 2).
134. Explain what is meant by questionable and legitimate
transformations.
136. Explain a method of finding approximately the roots of a
cubic equation.
136. Find the equation of the straight line passing through the
points of intersection oi x^ + y' + 2x + 4y — 11 — 0 and
X- + y^ - 2x - 2y = 0.
§124] PROGRESSIONS 233
137. What are the equations of the coordinate axes?
138. What is the locus of x^ = 4? of y' = 4? of a^ = 2/«?
of o2a;2 = 62!/2?
140. Solve { „ , .", „ .
141. Define series.
142. Define arithmetical progression; geometrical progression; har-
monical progression.
143. Define arithmetical mean; geometrical mean.
144. Derive formulas for I and s of an arithmetical progression.
146. Derive formulas for I and s of a geometrical progression.
146. Define an infinite geometrical progression. •
147. Derive the formula for the sum of an infinite geometrical
progression.
148. Find the value of 0.273273273 . .
149. A debt of $10,000 is to be paid in ten years. An equal amount
is paid at the end of each year. Find this amount if the indebtedness
draws interest at 5 percent.
150. An equal amount of money is deposited at the end of each year
for twenty years as a sinking fund to replace a piece of machinery
valued at $10,000. How much must be deposited at the end of each
year, if the deposits draw 4 percent compound interest.
CHAPTER IX
THE LOGARITHMIC AND THE EXPONENTIAL
FUNCTIONS
125. Historical Development. The almost miraculous power
of modern calculation is due, in large part, to the invention of
logarithms in the first quarter of the seventeenth century by a
Scotchman, John Napier, Baron of Merchiston. This invention
was founded on a very simple and obvious principle, that had
been quite overlooked by mathematicians for many genera-
tions. Napier'sinventionmay be explained as follows:^ Let there
be an arithmetical and a geometrical progression which are to be
associated together, as, for example, the following :
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
Now the product of any two numbers of the second line may be
found by adding the two numbers of the first progression above
them, finding this sum in the first Une, and finally taking the num-
ber lying under it ; this^Iatter number is the product sought. Thus,
suppose the product of 8 by 32 is desired. Over these numbers
of the second line stand the numbers 3 and 5, whose sum is 8.
Under 8 is found 256, the product desired. Now since but a
limited variety of numbers is offered in this table, it would be
useless in the actual practice of multiplication, for the reason
that the particular numbers whose product is desired would
probably not be found in the second line. The overcoming
of this obvious obstacle constitutes the novelty of Napier's inven-
tion. Napier proposed to insert any number of intermediate
terms in each progression. Thus, instead of the portion
0, 1, 2, 3, 4
1, 2, 4, 8, 16
1 Merely the fundamental principles of the invention, not historical details, are
given in what follows. For a very brief course in logarithms, only §§131-144
need be taken.
234
§126] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 236
of the two series we may wi-ite
0, I 1, U, 2, 2i 3, 3), 4
1, \/2, 2, Vs, 4, a/32, 8, ^128, 16
by inserting arithmetical means between the consecutive terms
of the arithmetical series and by inserting geometrical means
between the terms of the geometrical series. Let these be
computed to any desired degree of accuracy, say to two decimal
places. Then we have the series
A. P.
G.P.
0.0
1.00
0.5
1.41
1.0
2.00
1.5
2.83
2.0
4.00
2.5
5.66
3.0
8.00
Again inserting arithmetical and geometrical means between the
terms of the respective series we have:
A. P.
G.P.
0.00
1.00
0.25
1.19
0.50
1.41
0.75
1.69
1.00
2.00
1.25
2.38
1.50
2.83
1.75
3.36
2.00
4.00
2.25
4.76
By continuing this process each consecutive three figure number
may finally be made to appear in the second column, so that, to
this degree of accuracy, the product of any two such numbers
may be found by the process previously explained. The decimal
points of the factors may be ignored in this work, as for example,
the product of 2.38 X 14.1 is the same as that of 238 X 14.1
236 ELEMENTARY MATHEMATICAL ANALYSIS [§126 .
except in the position of the decimal point. The correct position
of the decimal point can be determined by inanection after the
significant figures of the product have been obtained. Using
the above table we find 2.38 X 14.1 = 33.6.
The above table, when properly extended, is a table of loga-
rithms. As geometrical and arithmetical progressions different
from those given above might havo been used, the number of
possible systems of logarithms is indefinitely great. The first
column of figures contains the logarithms of the numbers that
stand opposite them in the second column. Napier, by this
process, said he divided the ratio of 1.00 to 2.00 into "100 equal
ratios," by which he referred to the insertion of 100 geometrical
means between 1.00 and 2.00. The "number of the ratio" he
called the logarithm of the number, for example, 0.75 opposite
1.69, is the logarithm of 1.69. The word logarithm is from two
Greek words meaning " The number of the ratios." In order to
produce a table of logarithms it was merely necessary to compute
numerous geometrical means; that is, no operations except multi-
plication and the extraction of square roots were required. But
the numerical work was carried out by Napier to so many decimal
places that the computation was exceedingly difficult.
The news of the remarkable invention of logarithms induced
Henry Briggs, professor at Gresham College, London, to visit
Napier in 1615. It was on this visit that Briggs suggested the ad-
vantages of a system of logarithms ia which the logarithm of
10 should be 1, for then it would only be necessary to insert a
sufficient number of geometrical means between 1 and 10 to
get the logarithm of any desired number. With the encourage-
ment of Napier, Briggs undertook the computation, and in 1617,
published the logarithms of numbers from 1 to 1000 and, in
1624, the logarithms'of numbers from 1 to 20,000, and from 90,000
to 100,000 to fourteen decimal places. The gap between 20,000
and 90,000 was filled by a Hollander, Adrian Vlacq, whose table,
published in 1628, is the source from which nearly all the tables
since published have been derived.
126. Graphical Computation of the Terms of a Geometrical
Progression. Draw the lines y = x and y = rx, Fig. 100. From
the point (1, r) on ?/ = rx draw a horizontal line io y = x, thence
§127] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 237
a vertical line toy = rx, etc., thereby forming the "stairway" of
line segments between y = x and y = rx a,s shown in Fig. 100.
Then the points, N, P, Q, etc., have the ordinates r, r'\ r^, etc.,
as required, for, to obtain the ordinate of P, or PD, the value of x
used was OD = r, hence P is the point on y = rx for a; = r, or
y
PD
Likewise Q is by construction the point on y
rx for X = r^, hence the y of the point Q = r X r' = r^, etc.
8
T
6
Y
U
/ h
/ 1
4
q/
:/i
7
3
./
'//
/•
2
N
/
^
/
>r'
M
/
>■>■-
---'
■ r
r— "
4— ,-
]rr-
r^
J
c
a-2-1012345
Fig. 100.^ — Graphical construction of the successive terms of a G. P.
In the diagram r =3/2, and the curve isy = (3/2)*. /
The points P, Q, etc., are now carried horizontally to points
whose abscissas are, respectively, 2, 3, 4, etc., thus giving points
on the curve for y = r*.
127. Graphical Computation of Logarithms. In Fig. 100 the
termis of a geometrical progression of first term 1 and ratio IN = r
are represented as ordinates arranged at equal intervals along OX.
Fig. 100 is drawn to scale for the value of r = 1.5. Fig. 101 is
a similar figure drawn for r = 2, in which a process is used for
locating intermediate points of the curve, so that the locus may
238 ELEMENTARY MATHEMATICAL ANALYSIS [§127
be sketched with greater accuracy. The lines y =• x and y = rx
(in this case y = 2x) are drawn, and the "stairway" constructed
as before (See §126). Vertical lines drawn through a; = —2, —1,
0, 1, 2, 3, . . . and horizontal lines drawn through the hori-
zontal tread of each step of the stairWay divides the plane into a
large number of rectangles. Starting at M and sketching the
diagonals of successive cornering rectangles the smooth curve
-. -1 0 1 2 3 - .
Fig. 101. — Graphical construction of the curve y = 2".
MNP is obtained. Intermediate points of the curve are located by
doubling the number of vertical lines by bisecting the distances
between each original pair, and then by increasing the number of
horizontal lines in the following manner: Draw the line y = s/r x
(in the case of the Fig. 101, y = V'2 x). At the points where
this line cuts the vertical risers of each step of the "stairway"
(some of these points are marked .A, B, C in the diagram) draw a
§127] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 239
new set of horizontal lines. Each of the original rectangles is thus
divided into four smaller rectangles. Starting at M and sketching
a smooth curve along the diagonals of successive cornering rec-
tangles, the desired graph is obtained. By the use of the straight
line y = -s/r x another set of intermediate points may be located,
and so on, and the resulting curve thus drawn to any degree of
accuracy required. In explaining this process, the student will
show that the method of construction just used consists in the
doubling of the number of horizontal lines of the figure by the
successive insertion of geometrical means between the terms of a
geometrical progression, while at the same time the number of
vertical lines is successively doubled by the insertion of arithmet-
ical means between the terms of an arithmetical series. Thus the
graphical work of construction of the curve corresponds to the
successive insertion of geometrical and arithmetical means in the
two series discussed in §125.
As explained above, the ordinate y of any point of the curve
MNP of Fig. 101 is a term of a geometrical progression, and the
abscissa x of the same point is the corresponding term of an
arithmetical progression. Since, when y is given, the value of x
is determined, we say, by definition, that a; is a function of y
(§6). This particular functional relation is so important that
it is given a special name: x is called the logarithm of y, and the
statement is abbreviated by writing
X = logy,
but to distinguish from the case in which some other geometrical
progression might have been used, the ratio of the progression
may be written as a subscript, thus
X = logr?/,
which is read "x is the logarithm of y to the base r."
The ratio of the geometrical progression, or r, is called the base.
If we assume that the process of locating the successive sets of
intermediate points by the construction of successive geometrical
means will lead, if continued indefinitely, to the generation of
the curve MNP without breaks or gaps, then we may say that in
the equation
X = lOgry, ; (1)
240 ELEMENTARY MATHEMATICAL ANALYSIS [§127
the logarithm is a function of y defined for all -positive values of y
and for all halites of x.
It is seen at once from the method of construction used in Fig.
101 that the values of y at a; = 1, 2, 3, 4, ... , are respectively
y = r, T^, r-', r*, . . , and the values oi y a.t x = 1/2, 3/2, 5/2,
. . . , are y = r^, r^, r^, . . . , respectively, and similarly for
other intermediate values of x. In other words, the equation
connecting the two variables x and y may be written
y = r^ (2)
Thus, when the values of a variable x run over an arithmetical
progression {of first term 0) while the corresponding values of a
variable y run over a geometrical progression {of first term 1), the
relation between the variables may be written in either of the forms
(1) or (2) above. Equation (2) is called an exponential equation
and y is said to be an exponential function of x, while in (1) x
is said to be a logarithmic function of y. The student has fre-
quently been called upon in mathematics to express relations
between variables in two different or "inverse" forms, analogous
to the two forms y = r' and x = logr?/. For example, he has
written either
y = x^, 01 X = ± y/y;
and either
y = X ^ 01 X = y^
The graph of a function is of course the same whether the equation
be solved for x or solved for y.
Exercises
1. Write the following equations in logarithmic form:
(a) y = l(y; (d) u = 5';
(b) y = 3'; (e) z = o"; 1
(c) y = a^; if) u = 1.1'.
> As a matter of fact, both the arithmetical and the geometrical methods given
above define the function only tor rational values of x; that is, the only values of
X that come into view in the process explained above are whole numbers and
intermediate rational fractions like 2|, 2j, 2f, 2^j, 2j|, . . .
§128] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 241
2. Write the following equations in exponential form :
(o) X = logio y; (d) u = log, y;
(b) X = logs y, (e) t = logs z;
(c) X = logs y; (/) u = loga x.
3. Find the values of the following :
(a) logio 100; (d) log2 64;
(6) logs 25; (e) log, 81;
(c) logs 27; if) log ^16.
128. The Subtangent of the Exponential Curve. The
student is expected to construct the curves described in the
following exercises by the method of §127. The inch or 2 cm.
may be adopted as the unit of measure; the curves should be drawn
on plain paper within the interval from x = — 2toa;=-j-;2.
If tangents' be drawn to the curves at x = — 2, — 1, 0, 1, 2,
it will be noted, as nearly as can be determined by experiment,
that the several tangents to any one curve cut the X-axis at the
same constant distance to the left of the ordinate of the point
of tangency. This distance is called the subtangent of the curve.
This distance is greater than unity if r = 2 and less than unity
if r = 3. The value of the base for which the subtangent is exactly
unity is later shown to be a certain irrational or incommensurable
number, approximately 2.7183 . , represented in mathematics
1 It is not easy to draw accurately the tangent to a curve at a given point.
A number of instruments have been designed to assist in drawing tangents to
curves. One of these, called a "Radiator," will be found listed in most catalogs of
drawing instruments. Another instrument consists of a straight edge provided
with a vertical mirror as shown in Fig. 102. When the straight edge is placed across
Fig. 102. — Mirrored ruler for drawing the normal (and hence the '
tangent) to any curve.
a curve the reflection of the curve in the mirror and the curve itself can both bii
seen and usually the curve and image meet to form a cusp or angle. The straight
edge may be turned, however, until the image forms a smooth continuation of the
given curve. In this position the straight-edge is perpendicular to the tangent and
the tangent can then be accurately drawn. See Gramberg, Technische Messungen,
1911.
16
242 ELEMENTARY MATHEMATICAL ANALYSIS [§129
by the letter e, and called the Naperian base. This number, and
the number ir, are two of the most important and fundamental
constants of mathematics.
Exercises
Draw the following curves on plain paper using 1 inch or 2 cm. as the
unit of measure; make the tests referred to in the second paragraph of
§128.
1. Construct a curve similar to Fig. 101, representing the equation
X = log2 y, from a;=— 2toa; = +2, and draw tangents at a; = — 1,
X = 0, X = 1, X = 2. >
2. Construct the curve whose equation is a; = logs y from a; = — 2
to a; = +2, and draw tangents at a; = — 1, x = 0, a; = 1, x = 2.
3. Construct the curve whose equation is x = logs.? y, and show by
trial or experiment that the tangent to the curve at x = 2 cuts the X-
axis at nearly x = 1, that the tangent at x = 1 cuts the X^xis at
nearly x = 0, that the tangent at x = 0
cuts the X-axis at nearly x = — 1, etc.
4. Draw the curve x = logo.s y and
show that it is the same as the reflection
of X = log2 y in the mirror x = 0.
Note: The student must remember
that the experimental testing of the
properties of the tangents to the curves
called for above does not constitute mathe-
FiQ. 103. matical proof of the usual deductive sort
famUiar to him. The experimental tests
have value, however, in preparing the student for a rigorous in-
V estigation of these same properties when taken up in the calculus.
129. Slope of the Exponential Curve. Let MP, Fig. 103, be
any exponential curve, y = r. By the slope of the curve at P
we mean the slope of the tangent TP at P. We have just shown
experimentally that the length of the subtangent TD is constant
for all positions of the point P on the curve y = r". We can
then write
slope of curve at P = ^^^ = j> (1)
1 U K
where k is the constant length of TD. We can also write
slope of curve at P = cy, (2)
where c = t •
k
§130] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 243
From (2) we can conclude that: The slope of an exponential
curve at a given point is proportional to the ordinate at that point.
At the point (0, 1) the slope is c = t- As we have seen, the
value oik {= TD) depends on the value of r in the equation of the
curve, y = r". For some values of r it is less than 1, for others
greater than 1. We have defined e as that value of r for which
k = TD = 1. This is equivalent to defining e as that value
of r for which the curve y = r" has the slope unity at the point
(0, 1) . Later we shall adopt this definition of e.
Since c = t = 1 for the curve y = e", it follows from (2) that
for this particular curve of the family of curves y = r", the slope
at any point is equal to the ordinate.
The reasoning of this section is based on the experimentally de-
termined result that for a given exponential curve the subtangent
is of constant length.
130.* The Exponential Function. The expression a", where a
is any positive number except 1, has a definite meaning and
value for all positive or negative rational values of x, for the
meaning of numbers affected by positive or negative fractional
exponents has been fully explained in elementary algebra. The
process outlined above likewise defines logrS for aU rational
values of x, but not for irrational values of x, such as -\/2, VB, etc.
As a matter of fact the expression a' has, as yet, no meaning
assigned to it for irrational values of x; thus 10^^ has no meaning
by the definitions of exponents previously given, for \/2 is not a
whole number, hence 10"^^ does not mean that 10 is repeated
as a factor a certain number of times; also \/2 is not a fraction,
so that 10"^^ cannot mean a power of a root of 10. But if any
one of the numbers of the following sequence :
1, 1.4, 1.41, 1.414, 1.4142, 1.41421, . .
be used as the exponent of 10, the resulting power can be com-
puted to any desired number of decimal places. For example,
IQi" is the 141th power of the 100th root of 10; to find the 100th
root we may take the square root of 10, find the square root of
244 ELEMENTARY MATHEMATICAL ANALYSIS [§131
this result, then find its 5th root, finally finding the 5th root
of this last result.
If the various powers be thus computed to seven places we find:
10'* =25.11887
10'" =25.70396
101" « =25.94179
101.4142 =25.95374
101.41421 = 25.95434 .
101.414213 = 25.95452 . . .
101.4142135 = 25.95455 .
Now the sequence of exponents used in the first column is
found by extracting the square root of 2 to successive decimal
places. The sequence in the second column approaches a limit.
This limit is taken hy definition as the value of 10'^''.
In general, if x is an irrational number, a" is defined as the limit
of a sequence of numbers, o^', a'^', . . . , a^. . . , the exponents
xi, xi, . ., Xn,. . . being a sequence of rational mumbers
approaching a; as a limit.
It thus appears that if a and y are any given positive numbers,
there is a number x, rational or irrational, which satisfies the equa-
tion a' = y. The expression a' is called the exponential func-
tion of X with "base a.
131. Definitions. — In the exponential equation a' = y:
The number a is called the base.
The number y is called the exponential function of x to the base
a, and is sometimes written y = expoX.
The number x is called the logarithm of y to the base ,a, and
is written x = logay- Thus in the equation a" = y, x may be
called either the exponent of a or the logarithm of y.
The two equations,
y = a'
X = logay,
express exactly the same relations between x and y; one equation
is solved for x, the other is solved for y. The graphs are identical,
just as the graphs oi y = x^ and x = ± \/y are identical.
132. Common Logarithms. In the equation 10"° = y, x is
§133] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 245
called the common logarithm of y. It is also called the Brigg's
logarithm of y. Thus, the common logarithm of any number is
the exponent of the power to which 10 must be raised to produce
the given number. Thus 2 is the common logarithm of 100,
since 10" = 100; likewise 1.3010 will be found to be the con^mon
logarithm of 20 correct to 4 decimal places, since lO^-'"" = 20.00.
133. Systems of Logarithms. If in the exponential equation
y = a', where a is any positive number except 1, different values
be assigned to y and the corresponding values of x be computed
and tabulated, the results constitute a system of logarithms.
The number of different possible systems is unlimited, as abeady
noted in §125. -As a matter of fact, however, only two systems
have been computed and tubulated; the natural, or Naperian, or
hyperboUc, system, whose base is the incommensurable number e,
approximately 2.7182818, and the common, or Brigg's, system,
whose base is 10. The letter e is set aside in mathematics to
stand for the base of the natural system.
Natural logarithms of all numbers from 1 to 20,000 have
been computed to 17 decimal places. The common logarithms
are usually printed in tables of 4, 5, 6, 7 or 8 decimal places.
It will be found later that the graphs of all logarithmic functions
of the form x = logo y can be made by stretching or by contract-
ing in the same fixed ratio the ordinates of any one of the logarith-
mic curves. That is, the logarithms of one system can be ob-
tained from those of another system by multiplying by a constant
factor. For this reason numerical tables in more than one
system of logarithms are unnecessary.
In the following pages the common logarithm of any number n
wiU be written log n, and not logu n; that is, the base is supposed
to be 10 unless otherwise designated; In x for logeS and Ig x for
logic X are also used.
Exercises
Write the following in logarithmic notation:
1. 103 = 1000. 6. e" = y.
2. 10-3 ^ 0.001. 7. 10»" = 1.7783.
3. 10» = 1. 8. lO»Mio = 2.
4. IP = 121. 9. oi = a.
5. 16«" = 2. 10. 10i°8io!' = y.
Express the following in exponential notation :
246 Ea^EMENTARY MATHEMATICAL ANALYSIS [§134
11. logu 4 = 0.6021. 16. log-^^iOO = I
12. log 10000 = 4. 17. logj7(l) = -ll.
13. log 0.0001 = - 4. 18. logic 10 = i-
14. logs 1024 = 10. 19. log 1 = [O.
16. log. o = 1. 20. logal = [o.
134. Graphical Table.
function defined by the two progressions whose use was suggested
In Fig. 104 is shown the graph of the
10
N
/
/
/
/
/
/
L
-i
^f
10
N
/
or
/
A
r =
10
/
/
y
/
/
/
y"
y
^
^
^
1
0
0
1
0
2
0
3
0
4
0
5
0
6
0
7
0
8_
0
9fl
[o
Fig. 104. — The curve L = logioiV.
by Briggs to Napier, and which are referred to in the last para-
graph of §125. By inserting means three times between 0
and 1 in the arithmetical progression and between 1 and 10 in the
geometrical progression, we get-
A. P. or
logarithms
G. P. or
numbers
Exponential
form of G. P.
0.000
1.000
lOO-oO"
0.125
1.334
X00.126
0.250
1.778
100.260
0.375
2.371
100.375
0.500
3.162
IQO.eoo
0.625
4.217
100. 6S6
0.750
5.623
100.760
0.875
7.499
100.»76
1.000
10.000
IQi.ooa
§135] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 247
If we let L stand for the logarithm of the number N, the
functional relation is obviously L = logioiV, or iV = 10^. The
curve, Fig. 104, may now be used as a graphical table of logarithms
from which the results can be read to about 2 decimal places.
The logarithms of numbers between 1 and 10 may be read directly
from the graph. Thus, logio 7.24 = 0.860. If the logarithm is
between 0 and 1, the number is read directly from the graph.
Thus if the logarithm is 0.273, the number is 1.87.
If we multiply the readings of the A/^-scale by 10", we must add
n to the readings on the L-scale, for lO^A'' = 10^+".
If we divide the readings on the A''-scale by 10", we must
subtract n from the readings on the L-scale, for N/10" = 10^ ~".
This fact enables us to read the logarithms of all numbers from
the graph, and conversely to find the number corresponding to
any logarithm. Thus we have, log 72.4 = 1.860, log 724 =
2.860, log 0.724 = 0.860 - 1, log 0.0724 = 0.860 - 2.
If the logarithm is 1.273, the number is 18.7.
If'the logarithm is 2.273, the number is 187.
If the logarithm is 0.273 - 1, the number is 0.187.
If the logarithm is 0.273 - 2, the number is 0.0187.
We observe that the computation of a three place table -oi
logarithms would not involve a large amount of work. Such a
table has actually been computed in drawing the curve of Fig.
104. The original tables of Briggs and Vlacq involved an enor-
mous expenditure of labor and extraordinary skill, or even genius
in computation, because the results were given to fourteen places
of decimals.
135. Properties of Logarithms. The following properties of
logarithms follow at once from the general properties or laws of
exponents.
(1) The logarithm of 1 is 0 in all systems. For a" = 1, that
is, logal = 0. In Fig. 101, note that the curve passes through
(0, 1).
(2) The logarithm of the base itself in any system is 1. For
a^ = o, that is, log„a = 1. In Fig. 101, by construction N is
always the point (1, r), where r is the ratio of the first or funda-
mental progression in which means are inserted; in the present
notation, this is the point (1, a).
248 ELEMENTARY MATHEMATICAL ANALYSIS [§136
(3) Negative numbers have no logarithms. This follows at
once from Fig. 101. In Figs. 100, 101, and 104, note that the
curves do iiot extend below the X-axis.
Note: While negative numbers have no logarithms, this does not
prevent the computation of expressions containing negative factors
and divisors. Thus to compute (287) X (- 374), find (287) X (374)
by logarithms and give proper sign to the result.
136. Logarithm of a Product. Let n and r be any two positive
numbers, and let
logo n = X and logo r = y. (1)
Then, by definition of a logarithm, §131,
n — a' and r = a". (2)
Hence,
nr = a" a" = a''*^
Therefore, by definition of a logarithm, §131,
logo nr = X + y,
or, by (1),
log. nr = loga n + log. r. (3)
Hence, the logarithm of the product of two numbers is equal to
the sum of the logarithms of those numbers.
In the same way, if log. s = «, then
nrs = a"^-'-',
that is,
log. nrs = log. n + log. r + log. s.
Exercises
Find the results by the formulas and check by the curve of Fig. 104.
1. Given log 2 = 0.3010, and log 3 = 0.4771; find log 6; find log 18.
2. Given log 5 = 0.6990 and log 7 = 0.8451; find log 35.
3. Given log 9 = 0.9542, find log 81.
4. Given log 386 = 2.5866 and log 857 = 2.9330; find the logarithm
of their product.
6. Given log llx = 1.888 and log 11 = 1.0414; find log x.
137. Logarithm of a Quotient. Let n and r be any two positive
numbers, and let
log. n = X and log. r = y. (1)
§138] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 249
From (1), by the definition of a logarithm,
n = O' and r = a".
Hence,
n/r = a" -T- a" = a''".
Therefore, by definition of a logarithm,
loga {n'/r) = X - y,
or by (1)
logo (n/r) = logo n - log, r. (2)
Hence, the logarithm of the quotient of two numbers equals the
logarithm of the dividend minus the logarithm of the divisor.
Exercises
Find the results by the formulas and check by the curve of Fig. 104.
1. Given log 6 = 0.6990 and log 2 = 0.3010; find log (5/2); find
log 0.4.
2. Given log 63 = 1.7993, and log 9 = 0.9542; find log 7.
3. Given log 84 = 1.9243 and log 12 = 1.0792; find log 7.
4. Given log 1776 = 3.2494 and log 1912 = 3.2815; find log
1776/1912; find log 1912/1776.
5. Given log a;/12 = 0.4321 and log 12 = 1.0792, find log x.
138. Logarithm of any Power. Let n be any positive number
and let
logo n = X. (1)
From (1), by the definition of a logarithm,
n = a".
Raising both sides to the pth power, where p is any number what-
soever,
UP = a"'.
Therefore, by definition of a logarithm,
logo (w) = px,
or, by (1),
logo (n^) = p logon. (2)
Hence, the logarithm of any power of a number equals the logarithm
of the number multiplied by the index of the power.
250 ELEMENTARY MATHEMATICAL ANALYSIS [§139
The above includes the two cases: (1) the finding of the
logarithm of any integral power of a number, since, in this case
p is a positive integer; and (2) the finding of the logarithm of any
root of a number, since, in this case, p is the reciprocal of the index
of the root.
Exercises
1. Given log 2 = 0.3010; find log 1024; find log V2; find log y^.
2. Given log 1234 = 3.0913; find log Vl234; find log -^/i^Si.
3. Given log 5 = 0.6990; find log 53 ; find log sl.
4. Show that log (11/15) + log (490/297) - 2 log (7/9) = log 2.
6. Find an expression for the value of x from the equation 3' = 567.
Solution: Take the logarithm of each side;
X log 3 = log 567.
But
log 567 = log (3< X 7) = 4 log 3 + log 7.
Therefore
X log 3 = 4 log 3 + log 7,
or
X = 4 + (log 7)/(log 3).
6. Find an expression for x in the equation 5' = 375.
7. Given log 2 = 0.3010 and log 3 = 0.4771, find how many digits
in 6'°.
8. Find an expression for x from the equation
3» X 2»+i = -v/si^.
9. Prove that log (75/16) - 2 log (5/9) + log (32/243) = log 2.
139. Characteristic and Mantissa. The common logarithm
of a number is always written so that it consists of a positive
decimal part and an integral part which may be either positive
or negative. Thus log 0.02 = log 2 - log 100 = 0.3010 - 2.
Log 0.02 is never written — 1.6990.
When a logarithm of a number is thus arranged, special names
are given to each part. The positive or negative integral part is
called the characteristic of the logarithm. The •positive decimal
part is called the mantissa. Thus, in log 200 = 2.3010, 2 is
the characteristic and 3010 is the mantissa. In log 0.02 =
0.3010 — 2, (— 2) is the characteristic and 3010 is the mantissa.
Since log 1=0 and log 10 = 1, every number lying between 1
§139] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 251
an4 10 has for its common logarithm a number between 0 and 1 ;
that is, the characteristic is 0. Thus log 2 = 0.3010, log 9.99 =
0.9996, log 1.91 = 0.2810. Starting with the equation
log 1.91 = 0.2810,
we have, by §136,
log 19.1 = log 1.91 + log 10 = 0.2810 + 1 = 1.2810,
log 191 = log 1.91 + log 100 = 0.2810 + 2 = 2.2810,
log 1910 = log 1.91 + log 1000 = 0.2810 + 3 = 3.2810, etc.
Likewise, by §137,
log 0.191 = log 1.91 - log 10 = 0.2810 - 1,
log 0.0191 = log 1.91 - log 100 = 0.2810 - 2,
log 0.00191 = log 1.91 - log 1000 = 0.2810 - 3, etc.
Since thexharaoteristic of the common logarithm of any number
having its first significant figure in units place is zero, and since
moving the decimal point to the right or left is equivalent to
multiplying or dividing by a power of 10, or equivalent to adding
an integer to or subtracting an integer from the logarithm,
(§134): (1) the value of the characteristic is dependent merely
upon the position of the decimal point in the number; (2) the
value of the mantissa is the same for the logarithms of all
numbers that differ only in the position of the decimal point.
In particular, we derive therefrom the following rule for finding
the characteristic of the common logarithm of any number:
The characteristic of the common logarithm of a number equals
the number of places the first significant figure of the number is
removed from units' place, and is positive if the first significant
figure stands to the left of units' place and is negative if it stands
to the right of units' place.
Thus, in log 1910 = 3.2810, the first figure 1 is three places from
units' place and the characteristic is 3. In log 0.0191 = 0.2810
— 2, the first significant figure 1 is two places to the right of units'
place and the characteristic is — 2. A computer in determining
the characteristic of the logarithm of a number first points to
units' place and counts zero, then passes to the next place and
counts one and so on until the first significant figure is reached.
252 ELEMENTARY MATHEMATICAL ANALYSIS [§140
Logarithms with negative characteristics, Uke 0.3010 — 1,
0.3010 — 2, etc., should be written in the equivalent form
9.3010 - 10, 8.3010 - 10, etc.
Exercises ,
1. What numbers have 0 for the characteristic of their logarithm?
What numbers have 0 for the mantissa of their logarithms?
2. Find the characteristics of the logarithms of the following num-
bers: 1234, 5, 678, 910, 212, 57.45, 345.543, 7, 7.7, 0.7, 0.00000097,
0.00010097.
3. Given that log 31,416 = 4.4971, find the logarithms of the
foUowmg numbers: 314.16, 3.1416, 3,141,600, 0.031416, 0.31416,
0.00031416.
4. Given that log 746 = 2.8727, write the numbers which have the
foUowmg logarithms: 4.8727, 1.8727, 7.8727 - 10, 9.8727 - 10,
3.8727, 6.8727 - 10.
140. Logarithmic Tables. A table of common logarithms con-
tains only the mantissas of the logarithms of a certain convenient
sequence of numbers. For example, a four place table will con-
tain the mantissas of the logarithms of numbers from 100 to
1000; a five place table will usually contain the mantissas of
the logarithms of numbers from 1000 to 10,000, and so on. Of
course it is unnecessary to print decimal points or characteristics.
A table of logarithms should contain means for readily obtaining
the logarithms of numbers intermediate to those tabulated, by
means of tabular differences and proportional parts.'
The tabular differences are the differences between successive
mantissas. If any tabular difference be multiplied successively
by the numbers 0.1, 0.2, 0.3, . . . , 0.8, 0.9, the results are called
the proportional parts. Thus, from a four place table we find
log 263 = 2.4200. The tabular difference is given in the table
as 16. If we wish the logarithm of 263.7, the proportional part
0.7 X 16 or 11.2 is added to the mantissa, giving, to four places,
log 263.7 = 2.4211. This process is known as interpolation.
Corrections of this kind are made with great rapidity after a
1 The student is supposed to have Slichter's Four Place Tables, Macmillan A
Co., New York. The edition printed on three sheets of heavy manilla paper per-
forated to lit in notebook is preferred. See also tables at end of this book.
§140] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 253
little practice. It is obvious that the principle used in the
correction is the equivalent of a geometrical assumption that the
graph of the function is nearly straight between the successive
values of the argument given in the table. The corrections
should invariably be added mentally and all the work of interpolation
should be done mentally if the finding of the proportional parts by
mental work does not require multiplication beyond the range of
12 X 12. To make interpolations mentally is an essential practice,
if the student is to learn to compute by logarithms with any skiU
beyond the most rudimentary requirements.
A good method is the following: Suppose log 13.78 is required.
First write down the characteristic 1 ; then, with the table at your
left, find 137 in the number column and mark the corresponding
mantissa by placing your thumb above it or your first finger
below it. Do not read this mantissa, but read the tabular differ-
ence, 32. From the p. p. table find the correction, 26, for 8. Now
return to the mantissa marked by your finger, and read it increased
by 26, i.e., 1393; then place 1393 after the characteristic 1
previously written down.
The accuracy required for nearly aU engineering computations
does not exceed 3 or 4 significant figures. Four figure accuracy
means that the errors permitted do not exceed 1 percent of
1 percent. Only a small portion of the fundamental data of
science is reliable to this degree of accuracy.^ The usual meas-
urements of the testing laboratory fall far short of it. Only
in certain work "in geodesy, and in a few other special fields of
engineering, need more than four place logarithms be used.
Exercises
Knd the logarithms of the following :
1. 136. 4. 375.S 7. 2.758.
2. 752. 5. 217.6 8. 762,700.
3. 976. 6. 17.62 9. 0.1278.
^ Fundamental constants upon wMch much of the calculation in applied science
must be based are not often known to four figures. The mechanical equivalent of
heat is hardly known to 1 percent. The specific heat of superheated steam is even
less accurately known. The tensile, tortional, and compressive strength of no
structural material would be assumed to be known to a greater accuracy than the
above-named constants. Of course no calculated result can be more accurate than
the least accurate of the measurements upon which it depends.
254 ELEMENTARY MATHEMATICAL ANALYSIS [§U1
141. Anti-logarithms. If we wish to find the number which
has a given logarithm, it is convenient to have a table in which
the logarithm is printed before the number. Such a table is known
as a table of anti-logarithms. It is usually not best to print
tables of anti-logarithms to more than four places; to find a number
when a five place logarithm is given, it is preferable to use the
table of logarithms inversely, as the large number of pages required
for a table of anti-logarithms is a disadvantage that is not com-
pensated for by the additional convenience of such a table.
Exercises
From a four place table of anti-logarithms, find the numbers cor-
responding to the following logarithms:
1. 2.7864. 2. 3.1286. 3. 1.8152.
4. 9.6278 - 10. 5. 8.1278 - 10. . 6. 6.1785 - 10.
142. Cologarithms. Any computation involving multiplica-
tion, division, evolution, and involution may be performed by
the addition of a single column of logarithms. This possibility
is secured by using the cologarithms, instead of the logarithms, of
aU divisors. The cologarithm, or complementary logarithm,
of a number n is defined to be (10 — log n) — 10. The part
(10 — log n) can be taken from the table just as readily as log n,
by subtracting in order all the figures of the logarithm, including the
characteristic, from 9, except the last figure, which must be taken
from 10. The subtraction should, of course, be»done mentally.
Thus log 263 = 2.4200, whence colog 263 = 7.5800 - 10. In
like manner colog 0.0263 = 1.5800. It is obvious that the
addition of (10 — log n) — 10 is the same as the subtraction of
log n.
The convenience arising from this use may be illustrated as follows :
Suppose it is required to find x from the proportion
37.42 :x ::647 : v'0.S82!
We then have
2 log 37.4 = 3.1458
(1/2) log 0.582 = 9,8825 - 10
colog 647 = 7.1891 _ 10
log X = 0.2174
X = [1. 650].
§143] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 255
It is a good custom to enclose a computed result in square
brackets.
143. Arrangement of Work. All logarithmic work should be
arranged in a vertical column and should be done with pen and
ink. Study the formula in which numerical values are to be
substituted and decide upon an arrangement of your work in the
vertical column which will make the additions, subtractions, etc.,
of logarithms as systematic and easy as possible. Fill out the
vertical column with the names and values of the data before
turning to the table of logarithms. This is called blocking out
the work. The work is not properly blocked out unless every
entry in the work as laid out is carefully labelled, stating exactly
the name or value of the magnitude whose logarithm is taken,
and unless the computation sheet bears a formula or statement
fully explaining the purpose of the work.
Computation Sheet, Form M7, is suitable for general logarithmic
computation.
Illustration 1. Find the weight in pounds of a circular disk
of steel of radius 2.64 feet and thickness 0.824 inch, if the specific
gravity of the steel be 7.86.
Formula : Call r the radius in feet and t the thickness in inches.
Take 64.48 as the weight of one cubic foot of water. Then the weight
in pounds w is given by
w =-nrH</12)(64.48)(7.86).
Work:
logx
= 0.4971
21ogr
= 0.8432
log*
= 9.9159 - 10
colog 12
= 8.9208 - 10
log 64.48
= 1.8095
log 7.86
= 1.8954
log w
= 3.8819
w
= [761.9 lbs.]
Illustration 2. Compute the value of x if
■^ I O.K
X ==yj
1673 X 2.142_
3.871
256 ELEMENTARY MATHEMATICAL ANALYSIS [§143
Wobk:
log 0.1673 =9.2235-10
log 2,142 =0.3308
colog 3 . 871 = 9.4122 - 10
Sum = 8.9665 - 10
log a; =9.6555-10
X = [0.4524]
Remark 1. In writing a decimal fraction without an integral part,
always place a zero before the decimal point; thus 0. 1673.
Remark 2. Note the order of logarithmic work: First, write the
formula; next block out the work by writing down the first column, as
in illustrations above; finally hunt up logarithms from table and place
in second column of work.
Remark 3. After addition note that 23.8819 — 20 is written as
3.8819.
Remark 4. In dividing a logarithm like 8.9665 — 10 by 3, first
call the expression 28.9665 — 30 and then divide by 3. If division
by 5 had been required, the dividend would of course have been called
48.9665 - 50.
Exercises
1. Compute by logarithms the value of the following : 2.56 X 3.11
X 421; 7.04 X 0.21 X 0.0646; 3215 X 12.82 -^ 864.
2. Compute the following by logarithms: 81' ■^ 17«; 158\/0^;
(343/892)'; Vl893 \/l912/446='.
3. Compute the following by logarithms: (2.7182)'"'; (7.41)-*;
(8.31)«-".
4. Solve the following equations: 5* = 10; 3*-^ = 4; log» 71 =
1.21.
5. Find the amount of $550 for fifteen years at 5 percent compound
interest.
6. A corporation is to repay a loan of $200,000 by twenty equal
annual payments. How much will have to be paid each year, if
money be supposed to be worth 5 percent?
Let X be the amount paid each year. As the debt of $200,000 is
owed now, the present value of the twenty equal payments of x dollars
each must add up to the debt or $200,000. The sum of x dollars
to be paid n years hence has a present worth of only
X
(1.05)"
§143] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 267
if money be worth 6 percent compound interest. The present value,
then, of X dollars paid one year hence, x dollars paid two years hence,
and so on, is
1.05 (1.05)!' ' (1.05)3 , . . . , (105)20
This is a geometrical progression which can be summed by the usual
formula.
The result in this case is the value of an annuity payable at the end
of each year for twenty years that a present payment of $200,000 will
purchase. Four place tables will not give more than 3 place accuracy
in this and the following problem. To get 4 or 5 place accuracy,
6 place tables would be required.
7. It is estimated that a certain power plant costing $220,000 will
become entirely worthless except for a scrap value of $20,000 at the
end of twenty years. What annual sum must be set aside to amount
to the cost of replacement at the end of twenty years, if 5 percent
compound interest is realized on the money in the depreciation fund?
Let the annual amount set aside be x. In this case the twenty
equal payments are to have a value of $200,000 twenty years hence,
while in the preceding problem the payments were to be worth
$200,000 now. In this case, therefore,
a;(1.05)" + a;(1.05)" + x(1.05)" + . . .
+ x(1.05)2 + x(1.05) + s = $200,000.
The geometrical progression is to be summed and the resulting
equation solved for x.
8. The population of the United States in 1790 was 3,930,000 and
in 1910 it was 93,400,000. What was the average rate percent in-
crease for each decade of this period, assuming that the population
increased in geometrical progression with a uniform ratio for the entire
period.
9. Find the surface and the volume of a sphere whose radius is 7.211.
10. Find the weight of a cone of altitude 9.64 inches, the radius of
the base being 5.35 inches, if the cone is made of steel of specific
gravity 7.93.
11. Find the weight of a sphere of cast iron 14.2 inches in diameter,
if the specific gravity of the iron be 7.30.
12. In twenty-four hours of continuous pumping, a pump dis-
charges 450 gallons per minute; by how much will it raise the level of
water in a reservoir having a surface of 1 acre? (1 acre = 43560 sq.ft.)
17
258 ELEMENTARY MATHEMATICAL ANALYSIS [§144
144. Trigonometric Computations. Logarithms of the trigo-
nometric functions are used for computing the numerical value
of expressions containing trigonometric functions, and in the
solution of triangles. Right triangles, previously solved by
use of the natural functions, are often more readily solved by
means of logarithms. (See §62.) The tables of trigonometric
functions "contain adequate explanation of their use, so that
detailed instructions need not be given in this place. Two new
matters Of great importance are met with in the use of the loga-'
rithms of the trigonometric functions that do not arise in the use
of a table of logarithms of numbers, which, on that account, re-
quire especial attention from the student:
(1) In interpolating in a table of logarithms of trigonometric
functions, the corrections to the logarithms of all co-functions must
be svhtracted and not added. Failure to do this is the cause of
most of the errors made by the beginner.
(2) To secure proper relative accuracy in computation, the
S and T functions must be used in interpolating for the sine and
tangent of small angles.
In the following work, four place tables of logarithms are
supposed to be in the hands of the students.
Exercises
1. A lateral face of a right prism, whose base is a square 17.45 feet
on a side, is cut in a line parallel to the base by a plane making an
angle of 27° 15' with the face. Find the area of the section of the
prism made by the cutting plane.
2. The perimeter of a regular decagon is 24 feet. Find the area of
the decagon.
3. To find the distance between two points B and C on opposite
banks of a river, a distance CA is measured 300 feet, perpendicular
to CB. At A the angle CAB is found to be 47° 27'. Find the
distance CB.
4. In running a line 18 miles in a direction north, 2° 13.2' east,
how far in feet does one depart from a north and south line passing
through the place of beginning?
5. How far is Madison, Wisconsin, latitude 43° 5', from the earth's
axis of rotation, assuming that the earth is a sphere of radius 3960
miles?
§144] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 259
6. A man walking east 7° 15' north along a river notices that after
passing opposite a tree across the river he walks 107 paces before he
is in line with the shadow of the tree. Time of day, noon. How far
is it across the river?
7. Solve the right-angled triangle in which one leg = 2\/3 and the
hypotenuse = 2ir.
8. The moon's radius is 1081 miles. When nearest the earth, the
moon's apparent diameter (the angle subtended by the moon's disk as
seen from the position of the earth's center) is 32'.79. When farthest
from the earth, her apparent diameter is only 28'. 73. Find the
nearest and farthest distances of the moon in miles.
9. A pendulum 39 inches long vibrates 3° 5' each side of its mean
position. At the end of each swing, how far is the pendulum bob
above its lowest position?
10. If the deviation of the compass be 2° 1.14' east, how many feet
does magnetic north depart from true north in a distance of 1 mile
true north?
11. Solve
X : 1.72 = 427 : V2gh,
a g = 32.2 and h = 78.2.
12. A substance containing 20 percent of impurities is to be purified
by crystallization from a mother liquid. Each crystallization reduces
the impurity 88.6 percent. How many crystallizations will produce
a substance 0.9999 pure?
13. Compute the value of (1 — ae"'")" where a = 15.6, b = -t~'
\
X = 10, 71 = 2, 2/ = 2.5.
14. Find the volume of a cone if the angle at the apex be 15° 38'
and the altitude 17.48 inches.
15. The angle subtended by the sun's diameter as seen from the
earth is 32'.06. Find the diameter of the sun in miles, if the distance
from the earth to the sun be 92.8 million miles.
16. Compute by logarithms four values of p from the equation
V = 32.2(ii."8, for d = 2,3, 4, 5.
17. Solve 3' = 405 for the value of x.
18. Compute:
23.07 X 0.1354 X -s/234
13.54
What advantage is there in using the co-logarithm of the denomi-
nator?
260 ELEMENTARY MATHEMATICAL ANALYSIS [§145
145. * Logarithmic and Exponential Curves. The graphical
construction of the exponential curve has already been explained.
It was noted that curves whose equations are of the form y = r^
pass through the point (0, 1), and that the slope of the curves
for positive values of x is steeper the larger the value selected for
the number r. In a system of exponential curves y = r' passing
through the point (0, 1), or the point M of Fig. 105, we have
assumed (§129) that there is one curve passing through M
with slope 1. The equation of this particular curve we have
called y = e', thereby de-
fining the number e as that
value of r for which the
curve y = r' passes through
the point (0, 1) with slope 1.
In §130 there was de-
veloped on the basis of the
first definition of e, the
characteristic property of
the curve 2/ = e": The slope
of the curve y = e' at any
point is equal to the ordinate
of that point. This fact,
developed experimenrtaUy
in §129, will now be shown
to foUow necessarily from
the definition of e just
given.
Select the point P on the
curve y = e' at any point
desired. Draw a line through P cutting the curve at any neigh-
boring point Q. (Fig. 105.) A line like PQ that cuts a curve at
two points is called a secant line. As the point Q is taken nearer
and nearer to the point P (P remaining fixed), the limiting position
approached by the secant PQ is called the tangent to the curve
at the point P. This is the general definition of the tangent to
any curve.
The slope of the secant joining P to the neighboring point Q
is HQ/PH. As the point Q approaches P this ratio approaches
Fig. 105.-
-Definition of tangent to a
curve.
§145] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 261
the slope of the tangent to y = e' at the point P- Let OD = x
and PH = h; then OE = x + h, also DP = e' and EQ = e'+K
Since HQ is the y of the point Q minus the y of the point P, we
have
jjQ ^gx+h — gx^ ^e'^ — 1
Pff~ ^ ~ ^' /i
Now the slope of j/ = e* at P is the limit of the above expression
as Q approaches P, or as h approaches zero. That is
slope of e' at P = J^'J e' [^^] • (D
As the point Q approaches the point P, or as h approaches zero,
X does not change. Then
slope of e-^ at P = e- J^'J [^"T^] " ^2)
We now seek to find
limit ["e'^ - 11
h=Ol h J
if such limit exists.
Since the fraction (e* — l)/h does not contain x, its value, for
any value of h, and hence its hmit, is the same for every point of
the curve. If its value is calculated for any particular point of the
curve, as the point M, it will have this value at any other point as
P. .From equation (2) the slope of the tangent line at the point
M is
„ limitre^-1]
^ ^=0L h J
or
limit fe'- - 1]
h=Ol h i'
But by the definition of e, the slope of ?/ = e* at M is 1 .
S[^]-'- «>
Substituting this result in equation (2), we have
slope at P = e'. (4)
262 ELEMENTARY MATHEMATICAL ANALYSIS [§146
This expresses the fact that the slope oi y = e' at any point is e",
or is the ordinate y of that point, a fact that was first indicated
experimentally in §129.
Later an approximate value, 2.7183, will be found for e.
146. Comparison of the Curves y. = v and y = e*. In Fig.
105 the slope oiy = e' atP is given by DP measured by the unit
OM. The distance TD, the subtangent, is constant for all posi-
tions of the point P- We shall prove two theorems.
1. The curve for y = r" can be made from y = e' by multiplying
all of the abscissae of the latter by a constant. There is a number m
such that e"* = r'. Hence
y = r' may be written
y = (e"')' = e""'. Now
this curve is made from
2/ = e* by substituting mx
for X, or by multiplying
all of the abscissas of
y = e' hy 1/m.
2. The slope ofy = r' at
any point is a constant times
the ordinate of that point.
The curve y = r" can be
made from y = e' hy mul-
tiplying all of the abscissas
of the latter by 1/m.
Therefore the side TD of
the triangle PDT va. Fig.
105 will be multiplied by
1/m, the other side DP
remaining the same.
T)
o
IS
1
17
7\
1
«
a
u
1^
H
s
3
14
B
IS
B
A
n
11
in
1
7
-n
s
/
1
4
/
\
3
/
H
= log. X 1
-
-^
—
^
x=ey
.^
-4
-3
-2
-1
-1
/
!fs
?
3
4
5
6
7
r
9
0
1112 .3
14
-9,
f
^
- 2/.log, *|
-S
—
_
^
-4
^'
e'
-el
Fig. 106. — Exponential and logarith-
mic curves to the natural base e = log,, x
2.7183.
Hence the slope of the curve, or DP/TD, will be multiplied by
m, since the denominator of this fraction is multiplied by 1/m.
Hence, the slope oiy = r' at any point is m times the ordinate of
that point, where m satisfies the equation e" = r.
The curve y = e-' is (See §25) the curve y = e'' reflected in
the y-axis. This curve, as well as the curve y = log, x and its
symmetrical curve, are shown in Fig. 106. Sometimes the curve
y = e" Ss. called the exponential curve and the curve y = log. x
§146] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 263
is called the logarithmic curve. This distinction, however, has
little utility, as the equation of either locus can be expressed in
either notation.
The notation y = In x is often used to indicate the natural loga-
rithm of X and the notation y = Ig x, or y = log x, is used to stand
for the common logarithm of x.
Table IV.
The following table of powers of e is useful in sketching exponen-
tial curves.
eo.2 = 1.2214
ei^ -
= 1
6487
e-0.2
= 0.8187
e»-^ = 1.4918
eM =
= 1
3956
g-0,4
= 0.6703
e»« = 1.8221
e!-4 =
= 1
2840
g-0.6
= 0.5488
e»' = 2,2256
g-0.8
= 0.4493
e = 2.7183
e-'
= 0.3679
e^ = 7.3891
e-'-
= 0.1353
e' = 20.0855
e-3
= 0.0498
e* = 54.5982
e-"
= 0.0183
\
/
\
\\
.
/
/
?n= 0
^
fc:
m = 0
w
%
^fe
-2-10 12
Fig. 107. — A family of exponentials, y = e"
Exercises
1. Draw the curve y = e'- + e~^. Show that y is an even function
of X, that is, that y does not change when the sign of x is changed.
2. Draw the curve y = e" — e'". Show that y is an odd fun "
264 ELEMENTARY MATHEMATICAL ANALYSIS [§147
tion of X, that is, that the function changes sign but not absolute value
when the sign of x is changed.
3. Draw the graphs oi y = e", and y = e~''.
4. Draw the graphs of y = e*/*, and y = e~'/'.
6. Compare the curves: y = e*/*, y = e*''*, y = e', y = e'".
6. Sketch the curves y = 1', y = 2', y = Z', y = i", y = 5', y = &',
y = 8',y = 10"^, from a;=-3toa;=+3.
147. Change of Base and Properties of the Exponential Curv&
Consider the curves y = e' and y = a', Fig. 108, where a > e.
For a given y = OH, the abscissas HP\ and ffiPz are log, y and
logaj/, respectively. It has been shown (§146) that the curve
y = a' can be obtained from the curve y = e* by multiplying the
abscissas of the latter curve by — , where m is the number such that
■ a. (1)
T
"/
* ,
r
»/
"/
4/
y
B
*
2=^
Jt>
.yp
^
/>
7^P\
0
In other words.
That is,
EPi = - EP^.
m
log. y = - log' y-
(2)
■•^ As soon as m is known we have a
means of changing from a systeni of
^"*- I08--Co°iparison of logarithms with base e to one with base
V = er and y = a". ° .
a. The number — is called the modulus
m
of the logarithmic system whose base is a.
The modulus of the common system of logarithms is represented
by M. It is the value of — where m satisfies
e"> = 10, or TO = log, 10,
(3)
which is equation (1) for a = 10.
That is.
Hence,
e'^ = 10, or e = 10".
M = logio e = 0.4343.
(4)'
§147] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 265
Hence, if N represents any number,
logio N = 0.4343 log. N, (5)
log.N = — - — logioN = 2.3026 log, N. (6)
04343
From the definition of m and M,
100.«43 = g^ (7)
g2.3026 = 10. (8)
Incidentally it should be noted that, since M = —' from (3)
and (4),
^°S^°" = 1^' ^^^
A remarkable property of the logarithmic curve appears from
comparing the curves y = a' and y = a''*'''. The second of these
curves can be derived from y = a' by translating the latter curve
the distance h to the left. But y = a*+'' may be written y = a'^a^,
from which it can be seen that the new curve may also be
considered as derived from y = a' hy multipljang aU ordinates
oiy = a" by a*.
Translating the exponential curve in the negative x-direction is the
same as multiplying all ordinaies by a certain fixed number, or is
equivalent to a certain orthographic projection of the original curve
upon a plane through the X-axis.
Changing the sign of h changes the sense of the translation and
changes elongation to shortening or vice versa.
Exercises
1. Compare the curve y = e" with the curve y = 10*.
2. Graph the logarithmic spiral p = e>,6 being measured in radians.
Note : The radian measure in the margin of Form MZ should be
used for this purpose.
3. Graph p = e-«.
4. The pressure of the atmosphere is given in millimeters of mer-
cury by the formula
y = 760.e-»'/'i""'
where the altitude x is measured in meters above the sea level. Pro-
266 ELEMENTARY MATHEMATICAL ANALYSIS [§148
duce a table of pressure for the altitudes x =0; 10; 50; 100; 200;
300; 1000; 10,000; 100,000.
5. From the data of the last problem, find the approximate pressure
at an altitude of 25,000 feet.
6. Show that the relation of exercise 4 may be
written
X = 18,421 (log 760 - log y).
7. Determine the value of the quotient j for
the following values of x : 2, 3, 5, 7.
8. How large is e"""', approximately?
9. What is the approximate value of lO"""!?
148. Logarithmic Double Scale. The relation
between a number and its logarithm can be
shown by a double scale of the sort discussed in
§§3 and 10. Such a scale is shown in Fig. 109.
It may be constructed as follows: First con-
struct the uniform scale A, in which the unit
distance 0 — 1 is shown divided into 100 equal
parts. Opposite 0.3010 ( = log 2) of the A-
scale place a division line on the 5-scale marked
by the number 2. Opposite 0.4771 (= log 3)
of the A-scale place a division line of the B-
scale marked by the number 3. Likewise op-
posite 0.6021 (= log 4) of A place 4 on B; op-
posite 0.6990 (= log 5) of A place 5 on B; etc.
Intermediate points on B are similarly located —
for example the 2.1 mark on B should be
placed opposite 0.3222 (= log 2.1) on A.
The non-uniform scale B is called a loga-
rithmic scale, for the lengths measured along it
are proportional to the logarithms of the natural
numbers.
The double scale of Fig. 109 may obviously
be used as a table of logarithms. Thus from
it we may read log 7.1 = 0.85; log 3.3 = 0.52; log 1.5 = 0.175.
Since log 10a; = 1 + log x, it follows that, if the scales A and B,
Fig. 109, were extended another unit to the right, this second
s —
3
a
.a
a
ho
o
= i!
§149] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 267
unit would be identical to the first one, except in the attached
numbers. The numbers on the A-scale would be changed from
0.0, 0.1, 0.2, . .,1.0 to 1.0, 1.1, 1.2, . ,, 2.0, while those
on the non-uniform, or S-scale, would be changed from 1, 2, 3,
. . ., 10 to 10, 20,30, . ., 100.
Passing along this scale an integral number of unit intervals
corresponds thus to change of characteristic in the logarithms, and
to change in the position of the decimal point in the numbers. ■
It is not, however, necessary to construct more than one block of
this double scale, since we are at liberty to add an integer n to the num-
bers of the uniform scale, provided at the same time we multiply the
numbers of the non-uniform scale by 10". In this way we may obtain
any desired portion of the extended scale. Thus, we may change 0.1,
0.2, 0.3, . ., 1.0 on X to 3.1, 3.2, 3.3, . . ., 4.0, by adding 3 to
to each number, provided at the same time we change the numbers
1, 2, 3, 4, . ., 10 on the S-scale to 1000, 2000, 3000, 4000, . .,
10,000 by multiplying them by 10^. If n be negative (say — 2) we
may write, as in the case of logarithms, 8.0 — 10, 8.1 — 10, 8.2 —
10, . . ., 9.0 - 10, or, more simply, - 2, - 1.9, - 1.8, - 1.7,
., — 1.0, changing the numbers on the non-uniform scale at the
same time to 0.01, 0.02, 0.03, . ., 0.10.
Exercises
Read the following from the double scale. Fig. 109.
1. log 5.5 2. log 2.4 3. log 1.9 4. log 71
6. anti-log'O.74 6. anti-log 0.38 7. anti-log 1.38 8. anti-log 2.38
149. The Slide Rule. By far the most important apphcation
of the non-uniform scale ruled proportionally to log z, is the com-
puting device known as the slide rule. The principle upon which
the operation of the slide rule is based is very simple. If we have
two scales' divided proportionally to log x (A and B, Fig. 110),
so arranged that one scale may slide along the other, then slid-
ing one scale (called the slide) until its left end is opposite any
desired division of the first scale, selecting any desited division of
the slide, as at R, Fig. 110, and taking the reading of the original
scale beneath this point, as N, the product of, the two factors
whose logarithms are proportional to AB and BR can be read
268 ELEMENTARY MATHEMATICAL ANALYSIS [§149
X
directly from the lower scale at N. For AN is, by construction,
the sum oi AB and BR, and since the scales
were laid off proportionally to log x and marked
with the numbers of which the distances are the
logarithms, the process described adds the loga-
rithms mechanically, but indicates the results
in terms of the numbers themselves. By this
device all of the operations commonly carried
out by use of a logarithmic table may be per-
formed mechanically. Full description of the
use of the shde rule need not be given in de-
tail at this place, as complete instructions are
found in the pamphlet furnished with each
slide rule. A very brief amount of individual
instruction given to the student by the instruc-
tor will insure the rapid acquirement of skill in
the use of the instrument. In what follows,
the four scales of the slide rule are designated
from top to bottom of the rule, hy A, B, C, D,
respectively. The ends of the scales are called
the indices.
AH|Ordinary 10-inch sUde rule should give results
accurate to three significant figures, which is ac-
curate enough for most of the purposes of applied
science.
An exaggerated idea sometimes prevails con-
cerning the degree of accuracy required by work
in science or in applied science. Many of the
fundamental constants of science, upon which a
large number of other results depend, are known
only to three decimal places. In such cases
greater than three figure accuracy is impossible
even if desired. In other cases greater accuracy
is of no value even if possible. The real deside-
ratum in computed results is, first, to know by a
suitable check thai the work of compiUation is correct,
and, second, to know to what order or degree of ac-
curacy both the daia and the resuU are dependahle.
The absurdity of an undue number of decimal
^-2!* »
j3
a
§149] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 269
places in computation is illustrated by the original tables of loga-
rithms, which if now used would enable one to compute from the
radius of the earth, the circumference to 1/10,000 ■part o/ an inch.
The following matters should be emphasized in the use of the
slide rule :
(1) All numbers for the purpose of slide rule computation should
be considered as given with the first figure in units place. Thus
517 X 1910 X 0.024 should be considered as 5.17 X 1.19 X 2.4 X
10^ X 10' X 10~^ The result should then be mentally approxi-
mated (say 24,000) for the purpose of locating the decimal point,
and for checking the work.
(2) A proportion should always be solved by one setting of the
slide.
(3) A combined product and quotient like
aXhXcXd
■ rXsXt
should always be solved as follows:
Place runner on a of scale D;
set r of scale C to a of scale D;
runner to fe of C;
s of C to rurmer;
runner to c of C;
t of C to runner;
runner to d of C; find on D the significant figures of the '
result.
(4) The runner must be set on the first half of A for square
roots of numbers having an odd number of digits, and on the
second half of A for the square roots of other numbers.
(5) Use judgment so as to compute results in most accurate
manner — thus instead of computing 264/233, compute 31/233 and
hence find 264/233 = 1 + 31/233.^
(6) Besides checking by mental calculation as suggested in (1)
above, also check by computing several neighboring values and
graphing the results if necessary. Thus check 5.17 X 1.91 X 2.4
by computing both 5.20 X 19.2 X 2.42 and 5.10 X 1.90 X 2.38.
1 Show by trial that this* gives a more accurate result.
270 ELEMENTARY MATHEMATICAL ANALYSIS [§149
Exercises
Compute the following on the slide rule:
1. 3.12 X 2.24; 1.89 X 4.25; 2.88 X 3.16; 3.1 X 236.
2. 8.72/2.36; 4.58/2.36; 6.23/2.12; 10/3.14.
3. 32.5 X 72.5; 0.000116 X 0.00135; 0.0392/0.00114.
4. 3,967,000 H- 367,800,000. g 78.5 X 36.6 X 20.8
, 6.64X42.6 8.75X5.25 ' „ .^■'^^J^?^^
32.5 ' 32.3
Solve the proportion
6.46 X 57.5 X 8.55
3.26 X296 X 0.642'
X : 1.72 = 4.14 : V^gh.
where g = 32.2 andA^ = 78.2.
o n , VlTl X 1.41
9. Compute 166.7X4.5'
10. The following is an approximate formula for the area of a seg-
ment of a circle : <•
A = h'/2c + 2ch/3,
where c is the length of the chord and h is the altitude of the segment.
Test this formula for segments of a circle of unit radius, whose arcs
are 7r/3, ir/2, and tt radians, respectively.
11. Two steamers start at the same time from the same port; the
first sails at 12 miles an hour due south, and the second sails at 16
miles an hour due east. Knd the bearing of the &st steamer as seen
from the second {l) after one hour, (2) after two hours, and compute
their distances apart at each time.
The following exercises require the use of the data printed herewith.
An "acre-foot" means the quantity of water that would cover 1
acre 1 foot deep. "Second-foot" means a discharge at the rate of 1
cubic foot of water per second. By the "run-off" of any drainage
area is meant the quantity of water flowing therefrom in its surface
stream or river, during a year or other interval of time.
1 square mUe = 640 acres.
1 acre = 43,560 square feet.
1 day = 86,400 seconds.
1 second-foot = 2 acre-feet per day, approximately.
1 cubic foot = 7J gallons, approximately.
1 cubic foot water = 62^ pounds water, approximately.
1 h.p. = 550 foot-pounds per second.
450 gallons per minute = 1 second-foot, approximately.
§150] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 271
Each of the following problems should be handled on the slide rule as
a continuous piece of computation.
12. A drainage area of 710 square miles has an annual run-off of
120,000 acre-feet. The average annual rainfall is 27 inches. Find
what percent of the rainfall appears as run-off.
13. A centrifugal pump discharges 750 gallons per minute against
a total lift of 28 feet. Find the theoretical horse power required.
Also daily discharge in acre-feet if the pump operates fourteen hours
per day.
14. What is the theoretical horse power represented by a stream
discharging 550 second-feet if there be a fall of 42 feet?
15. A district containing 25,000 acres of irrigable land is to be sup-
plied with water by means of a canal. The average annual quantity
of water required is Sf feet on each acre. Find the capacity of the
canal in second-feet, if the quantity of water required is to be delivered
uniformly during an irrigation season of five months.
16. A municipal water supply amounts to 35,000,000 gallons per
twenty-four hours. Find the equivalent in cubic feet per second.
17. A single rainfall of 3.9 inches on a catchment area of 210 square
mUes is found to contribute 17,500 acre-feet of water to storage reser-
voir. The run-off is what percent of the rainfall in this case?
150. Semi-logarithmic Coordiaate Paper. Fig. Ill represents
a sheet of rectangular coordinate paper, on which ON has been
chosen as the unit of measure. Along the right-hand edge of this
sheet is constructed a logarithmic scale LM of the type discussed
in §148, i.e., any number, say 4, on the scale LM stands opposite
the logarithm of that number (in the case named opposite 0.6021)
on the uniform scale ON.
Let us agree always to designate by capital letters distances
measured on the uniform scales, and by lower case letters dis-
tances measured on the logarithmic scale. Thus Y will mean the
ordinate of a point as read on the scale ON, while y will mean the
ordinate of a point as read on the scale LM. Moreover, we agree
to plot a function, using logarithms of the values of the function
as ordinates and the natural values of the argument, or variable,
as abscissas.
Let PQ be any straight line on this paper, and let it be required
to find its equation, referred to the uniform a;-scale OL and the
logarithmic 2/-scale LM. We proceed as follows :
272 ELEMENTARY MATHEMATICAL ANALYSIS [§150
The equation of this line, referred to the uniform Z-axis OL
and the uniform Y-axis, ON, where 0 is the origin, is
Y = mx + B,
m being the slope of the line, and B its F-intercept. ,Now, for the
line PQ, m = 0.742 and B = 0.36, so that the equation of PQ is
Y = 0.742a: + 0.36. (1)
To find the equation of this curve referred to the scales LM and
OL, it is only necessary to notice that
y = log 2/
;v
Q
I
%
1.U
/^
/
9
.8
.7
.6
y
7
/
y
6
S
Y
/
y
y
3
<
i2
IL
.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0^
Fig. 111.— The theory of the use of semi-logarithmic paper.
so that we obtain
log y = 0.742a; + 0.36. (2)
The intercept 0.36 was read on the scale ON, and is therefore the
logarithm of the number corresponding to it on the scale LM.
That is, 0.36 = log 2.30. Substituting this value in equation
(2) we obtain
log V = 0.742a; + log 2.30,
§150] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 273
which may be written
log y - log 2.30 = 0.742a;,
log 2|o = 0-7*2..
}^M *=' ^
cS c
=■ ° o"' w-°
c
/
1,
/
.
/
8
/
y
7
y
0
y
'
6 =
/
b
y
/^
4
ny
/
S
/
•^
S
/
y
4
2
2
1
0.1
n
A L 0.1 0.2 0:3 0.4 0.5 o.< 0.7 0.8 0.0 \,0B
Semi LosarTthmlc Paper
Fig. 112, — Illustration of squared paper, form M5. The finer rulings
of form M5 have, however, been omitted.
Changing to exponential notation this becomes
2.30
10»'
y = 2.30(10°'«»). (3)
In general, if the equation of a straight line referred to the scales
OL and ON is
F = m + S, , (4)
18
274 ELEMENTARY MATHEMATICAL ANALYSIS [§150
its equation referred to the scales OL and LM may be obtained by
replacing Y by log y and B by log 6 in the manner described above,
giving
log y = mx + log 6, (5)
which, as above, may be reduced to the form
y = bio"*. (6)
This is the general equation of the exponential curve. Hence:
Any exponential curve can he represented by a straight line, provided
ordinates are read from a suitable logarithmic scale, and abscissas
are read from a uniform scale.
Fig. 112 represents the same line PQ, y = (2.30)10°'"', as
Fig. 111. The two figures differ only in one respect; in Fig. Ill
the rulings of the uniform scale ON are extended across the page,
while in Fig. 112 these rulings are replaced by those of the scale
LM.
Coordinate paper such as that represented by Fig. 112 is known
as semi-logarithmic paper. It affords a convenient coordinate
system for work with the exponential function.
Every point on PQ (Fig. 112) satisfies the exponential equation
y = 2.30(10" '^2-).
Thus, in the case of the point R,
3.98 = 2.30(10'''")'''2»
= 2.30(10»-238).
The slope of any line on the semi-logarithmic paper may be read
or determined by means of the uniform scales BC and AB oi form
M5. The scale AD of form M5 is the scale of the natural loga-
rithms, so that any equation of the form y = e"" can be graphed
at once by the use of this scale. Thus, the line y = e"''(Fig.
113) passes through the point A or (0, 1), and a point on BC op-
posite the point marked 1.0 on AD. Note that 1.0 on scale AD,
- 2.718 on the non-uniform scale of the main body of the paper,
and 0.4343 on the scale BC aU fall together, as they should.
To draw the line y = 10""', the corner D of the plate may be
taken as the point (0, 1). On the line drawn once across the sheet
representing y = 10"*, y has a range between 1 and 10 only.*
§150] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 275
To represent the range of y between 10 and 100, two or more sheets
of form M5 may be pasted together, or, preferably, the continua-
tion of the line may be shown on the same sheet by suitably chang-
ing the numbers attached to the scales AB and BC. Thus Fig.
113 shows in this manner y = IW".
Il 1 1 1 1 1 1 1 1
1,
,,,,
9\
/,*
\
/
/ !
8 \
/ I R
\
/
7
N,
/
/
/ '
S,
/
.
6
V
/
/
g
\,
1
/
/
5
\
1
/
/
6
V
/
/
1
4\
/
1
4
/
/
a
\
/
/
8
^
/
3
1
1
t
/
\
/
^
a
/
A
>>
■i
^
/
/
/
I
I
t
/
/
^
<
\
s
1
^
/
^
'^
1
1
1
1
1
1
1
1
\
\
\
.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 l.OB
Fig. 113. — Seini-logarithmic coordinate paper. The dotted line
gives two sections oi y= 10^"^.
Remember that on semi-logarithmic paper the line
y = bio""^ (7)
passes through the point (0, 6) with slope m. Note that
f = lo^C* - »)
(8)
passes through the point (a, 6) with slope m.
Illtjstration 1. Draw the curve ^x = log ^y on semi-logarithmic
paper.
This is the curve y = 2(105"^). This curve passes through the point
(0, 3) with slope |, hence can readily be drawn.
276 ELEMENTARY MATHEMATICAL ANALYSIS [§150 _
lUiTTBTBATiON 2. Draw the curve y = 3(10'^""*^].
From (8) above it is seen that this curve passes through the point
(2, 3) with slope 2.
Illustration 3. Plot the following data upon semi-logarithmic
paper and find, if possible, the equation connecting the x- and y-
values.
10
8
I
^
^
4
S
2
1
^
•
>
^
^
^
^
0.2 0.3
Fig. 114.-
0 4 0 5 0.6 0.7 0.8 as
-Diagram for Illustration 3.
X
y
0.2
3.18
0.4
3.96
0.6
5.00
0.8
6.30
The points plotted upon semi-logarithmic paper he on a straight
line as shown in Fig. 114. Hence, it is possible to find the equation
connecting x and y. The equation of this straight line is
Y = \x^r log 2.51,
§151] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 277
or
or
log V = hx -\- log 2.51,
'• y _ ^
log
2.51 2'
y
2.51
= 10'
V = 2.51(10»),
an empirical equation connecting the x- and j^-values of the table.
Exercises
On semi-logarithmic paper draw the following:
1. y = 10", y = W, y = 10»^.
2. y = 10-==, y = IQ-'i*, y = 10-»«.
3. y = e^', y = e".
i. y = e"", y = e~^'.
5. Sx = log y, (l/2)a; = log y.
6. Find an empirical equation connecting the x- and the y-values
given in the accompanying table.
X
y
0.2
0.4 ,
0.6
0.8
5.8
3.4
2.6
On semi-logarithmic paper draw the following:
^. y = 10'/2, y = io»/io.
■ 8. Graph y ^ 2(10)' and | = 10'-'.
161.* The Compound Interest Law. Computation of e. The
law expressed by the exponential curve was called by Lord Kelvin
the compound interest law and since that time this name has
been .generally used. It is recalled that the exponential curve
was drawn by using ordinates equal to the successive terms of
a geometrical progression which are uniformly spaced along the
278 ELEMENTARY MATHEMATICAL ANALYSIS [§151
X-axis. Since the amount of any sum at compound interest is
given by a term of a geometrical progression, it is obvious that a
sum at compound interest accumulates by the same law of growth
as is indicated by a set of uniformly spaced ordinates of an expo-
nential curve; hence the term "compound interest law," from
this superficial view, is appropriate. The detailed discussion
that follows will make this clear:
Let $1 be loaned at r percent per annum compound interest.
At the end of one year the amount is (1 -|- r/100).
At the end of two years the amount is (1 -|- r/100)'',
■ and at the end of t years it is (1 + r/100)'.
If interest be compounded semi-annually, instead of annually,
the amount at the end of t years is (1 -|- r/200)''',
and if compounded monthly the amount at the end of the same
period is (1 +-r/\2QQy^'
or if compounded n times per year y= {1 + r/lOOn)"',
where t is expressed in years. Now if we find the limit of this
expression as n is increased indefinitely, we will find the amount of
principle and interest on the hypothesis that the interest was
compounded conlinuously . For convenience let r/lOOn = 1/m.
Then
2/ = (1 + 1/m)-'/'»», (1)
where the limit is to be taken as m or n becomes infinite. Calling
(1 + l/uY = f(u) (2)
and expanding by the binomial theorem for any integral value
of u we obtain
r/- \ II fi / \ i w(w — 1) 1 ,
f{u) = 1 + u{l/u) + \2 ^ + • ■ •
= 1 -I- 1 + (1 - l/w)/2! -I- (1 - 1/m)(1 - 2/u)/3\ -I- ... (3)
In the calculus it is shown that the limit of this series as u becomes
infinite is the limit of the series
l-hl + l/2!-M/3!-h ... (4)
The limit of this series is easily found; it is, in fact, the Naperian
base e. It is shown in the calculus that the restriction that u
shall be an integer may be removed, so that the limit of (3) may
be found when m is a continuous variable.
§152] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 279
It is easy to see that the Umit of (4) is > 2^ and < 3. The sum
of the first three terms of the series (4) equals 2i; the rest of the
terms are positive, therefore e>2^. The terms of the series (4),
after the first three, are also observed to be less, term for term, than
the terms of the progression:
(1/2)2 + (1/2)3 + (5)
But this is a geometrical progression the limit of whose sum is 1/2.
Therefore (3) is always less than 2| + 5, or 3. The value of e is
readily approximated by the following computation of the first
8 terms of (4):
2.00000 = 1 + 1
0.50000 = 1/2!
0.16667 = 1/3!
0.04167 = 1/4!
0.00833 = 1/5!
0.00139 = 1/6!
0.00020 = 1/7!
Sum of 8 terms = 2.71826
The value of e here found is correct to four decimal places.
Returning to equation (1) above, the amount of $1 at r percent
compound interest compounded continuously is
y = e"/""- (6)
Thus $100 at 6 percent compound interest, compounded annually,
amounts, at the end of ten years, to
y = 100(1.06)"' = $179.10.
The amount of $100 compounded continuously for ten years is
y = 100e«-6= $182.20
The difference is thus $3.10
152. Logarithmic Increment. The compound interest law is
one of the important laws of nature. As previously noted, the
slope or rate of increase of the exponential function
y = ae'"
at any point is always proportional to the ordinate or to the value
280 ELEMENTARY MATHEMATICAL ANALYSIS [§152
of the function at that point. Thiis when in nature we find any
function or magnitude that increases at a rate proportional to itself
we have a case of the exponential or compound interest law.
The law is also frequently expressed by saying, as has been re-
peatedly stated in this book, that the first of two magnitudes varies
in geometrical progression while a second magnitude varies in arith-
metical progression. A famUiar example of this is the increased
friction as a rope is coiled around a post. A few turns of the
hawsers about the bitts at the wharf is sufficient to hold a large
ship, because as the number of turns increases 'In arithmetical
progression, the friction increases in geometrical progression.
Thus the following table gives the results of experiments to de-
termine what weight could be held up by a one-pound weight,
when a cord attached to the first weight passed over a round peg
the number of times shown in the first column of the table:
Average logarithmic increment =
n = number of
turns of the cord
on the peg
w = weight juBt held
in equilibrium by
one-pound weight
Logs of preceding
numbers
d = logarithmic
increment
1
1.6
3.0
5.1
8.0
14.0
23.0
0.204
0.477
0.708
0.903
1.146
1.362
1
li
2
21
3
0.273
0.231
0.195
0.243
0.216
0.23
If the weights sustained were exactly in geometrical progression,
their logarithms would be in arithmetical progression. The test
for this fact is to note whether the differences between logarithms
of successive values are constant. These differences are known
as logarithmic increments or in .case they are negative, as loga-
rithmic decrements. In the table the logarithmic increments
fluctuate about the mean value 0.23.
The equation connecting n and w is of the form
w = 10'"'"' or n = m log w
§153] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 281
By graphing columns 1 and 3 on squared paper, the value of m is
determined and we find
w = lO"-^'", or n = 2.2 log w.
Another way is to graph columns 1 and 2 on semi-logarithmic
paper.
An interesting example of the compound interest law is Weber's
law in psychology, which states that if stimuli are in geometrical
progression, the sense perceptions are in arithmetical progression.
163. Modulus of Decay, Logarithmic Decrement. In a very
large number of cases in nature a function obeying the "compound
interest" law appears as a decreasing function rather than as an
increasing function, so that the equation is of the form
y = ae~>"', (1)
where ( — 6) is essentially negative, b is the modulus of decay, or
the logarithmic decrement, corresponding to an increase of x
by unity. The following are examples of this law:
(1) If the thickness of panes of glass increase in arithmetical pro-
gression, the amount of light transmitted decreases in geometrical
progression. That is, the relation is of the form
L = oe-«, (2)
where t is the thickness of the glass or other absorbing material and L
is the intensity of the light transmitted. Since when t = 0 the light
transmitted must have its initial intensity, Lo, (2) becomes
L = Loe-«. (3)
The constant 6 must be determined from the data of the problem.
Thus, if a pane of glass one unit thick absorbs 2 percent of the incident
light,
U = 100, Z, = 98 for « = 1,
and 98 = 100e-»,
or log 98 - log 100 = - 6 log e.
Therefore 6 = j^j^ = 0.02
The light transmitted by ten panes of glass is then
iio = 100e-"'f»»«) = 100e-»-2,
282 ELEMENTARY MATHEMATICAL ANALYSIS [§153
or, by the table of §146,
Lio = 100/1.2214 = 82 percent
(2) Variation in atmospheric pressure with the altitude is usually
expressed by Halley's Law,
p = 760e-*/8»»»,
where h is the altitude in meters above sea level and p is the atmos-
pheric pressure in millimeters of mercury. See §147, Exercises 4, 5, 6.
(3) Newton's law of cooling states that a body surrounded by a
medium of constant temperature loses heat at a rate proportional
to the difference in temperature between it and the surrounding
medium. This, then, is a case of the compound interest law. If
6 denotes temperature of the cooling body above that of the surround-
ing medium at any time t, we must have
e = ae~K
The constant a must be the value of 8 when i = 0, or the initial tem-
perature of the body.
Exercises
1. A thermometer bulb initially at temperature 19°.3 C. is exposed to
the air and its temperature B observed to be 14°.2 C. at the end of
twenty seconds. If the law of cooling be given by e = ffoe"", where
t is the time in seconds, find the value of 6 and 6.
Soltjtion: The condition of the problem gives 9 = 19.3 when < = 0,
hence, Bo = 19.3. Also, 14.2 = 19.3e-206. This gives
log 19.3 - 20b log e = log 14.2,
from which 6 can be readily computed.
2. If IJ percent of the incident light is lost when Ught is directed
through a plate of glass 0.3 cm. thick, how much light would be lost in
penetrating a plate of glass 2 cm. thick? ■
3. Forty percent of the incident light is lost when passed through
a place of glass 2 inches thick. Find the value of a in the equation
L = LoB'"', where t is thickness of the plate in inches, L is the percent
of light transmitted, and Lo = 100.
4. As I descend a mountain the pressure of the air increases each
foot by the amount due to the weight of the layer of air 1 foot thick.
As the density of this layer is itself proportional to the pressure, show
that the pressure as I descend must increase by the compound inter-
est law.
6. Power is transmitted in a clock through a train of gear wheels
§154] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 283
n in number. If the loss of power in each pair of gears is 10 per-
cent, draw a curve showing the loss of power at the nth gear.
Note: The graphical method' of §121, Figs. 98, 99, may appro-
priately be used.
6. Given that the intensity of light is diminished 2 percent by
passing through one pane of glass, find the intensity / of the light
after passing through n panes.
7. The temperature of a body cooling according to Newton's law
.^fell from 30° to 18° in six minutes. Find the equation connecting
the temperature of the body and the time of cooling.
154. Empirical Curves on Semi-logarithmic Coordinate Paper.
One of the most important uses of semi-logarithmic paper is in
determining the functional relation between observed data, when
such data are connected by a relation of the exponential form
as already indicated in §160. Suppose, for example, that the
following are the results of an experiment to determine the law
connecting two variables x and y :
0.04 0.18 0.36 0.51 0.685 0.833 0.97
5.3 4.4 3.75 3.1 2.6 2.33 1.9
If the equation connecting x and y is of the exponential form, the
points whose coordinates are given by corresponding values of x
and y in the table will lie in a straight line on semi-logarithmic
paper, except for such errors as may be due to inaccuracies in the
observations. Plotting the points on semi-logarithmic coordinate
paper, we find that they lie nearly on the line PQ (Fig. 115).
Assuming that, if the data were exact, the points would lie exactly
on this line,' we may proceed to determine the equation of this line
as approximately representing the relation between x and y.
It is easy to find the equation of such a line referred to the uni-
^ We would not be at liberty to make such an assumption if tte variation of the
points away from the line was of a character similar to that represented by the dots
near the top of Fig. 115. These points, although not departing greatly from the
line shown; depart from it systematically. That is, they lie below it at each end ,
and above it in the center, seeming to approximate a curve (such as the one shown
dotted") more nearly than the line. The points arranged about the line PQ depart
as far from that line as do the points above the higher line, but they do not depart
systematically, as if tending to lie along a smooth curve. When points arrange
themselves as at the top of Fig. 115, one must infer that the relation connecting
the given data is not exponential in character.
284 ELEMENTARY MATHEMATICAL ANALYSIS [§154
form scale AB and BC of form M5. We may imagine that all
rulings are erased and replaced by extensions of the uniform AB
scale, as in Fig. 111. The equation of the line PQ is then
Y = mx + B, (1)
where m is the slope, and B is the T-intercept. Now, for PQ,
O-Jlf
f-.
iiiiiiiii
IIIIIIIII
rjmiiii
^Vs,
9:
s
8
"^v
8 :
N,
-
7
^<=r
,
^V
C
^*4
V
P
N,
*^^[) :
^^
4
^^^^
4 :
■~~^v^
1
3
;^
8 "
^
-^
1
^v
■^^2 =
1
1 :
"
""
A L »'' 0.2 0.3 0.4 O.S 0.6 0.7 0.6 0.0 1.0B
Semi Logarithmic Paper
Fig. 115. — Empirical equations determined by use of form Mb.
m = — 0.447 and B = 0.730 = log 5.37. Equation (1) becomes,
therefore
y = - 0.447a; + log 5.37
or, replacing Y by log y, in order to refer the curve to the
scales AB and LM,
log y - log 5.39 = - 0.447a:,
whence
y = s.ascio-""'*) (2)
§155] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 285
If it is desired to express the relation to the base e instead of
base 10, we may note 10 = e2-''026 (§147, equation (8)), or, sub-
stituting in (2),
y = 5.39 (e2-303)-o.447x
= 5.39 e-i»"« (3)
The same result could have been obtained directly by determin-
ing the slope of PQ from the uniform scale AD at the left of
Form Mb.
155. Change of Scale on Semi-logaritbmic Paper. A sheet of
semi-logarithmic paper, form Af5, is a square. If sheets of this
paper be arranged "checker-board fashion" over the plane, then
the vertical non-uniform scale will be a repetition of the scale LM,
Fig. 115, except that the successive segments of length LM must be
numbered 1, 2, 3, . ,9 for the original LM, then 10, 20,
30, , 90 for the next vertical segment of the checker-board,
then 100, 200, 300, , 900, for the next, etc. It is obvious,
therefore, that the initial point A of a sheet of semi-logarithmic
paper may be said to have the ordinate 1, or 10, or 100, etc., or
10~^, lO"'', etc., as may be most convenient for the particular
graph under consideration. The horizontal scale being a uniform
scale, any values of x may be plotted to any convenient scale on
it, as when using ordinary squared paper. However, if the hori-
zontal unit of length (the length AB, form Mb) be taken as any
value different from unity, then the slope m of the line PQ drawn
on the semi-logarithmic paper can only be found by dividing its
apparent slope by the scale value of the side AB. That is, the
correct value of m in
y — &10""'
is, in all cases,
_ apparent slope of PQ
scale value of AB
The "apparent slope" of PQ is to be measured by applying any
convenient uniform scale of inches, centimeters, etc., to the
horizontal and vertical sides of a right triangle of which PQ is the
hypotenuse.
286 ELEMENTARY MATHEMATICAL ANALYSIS [§156
Exercises
1. A thermometer bulb initially at temperature 19°.3 C. is exposed
to the air and its temperature e noted at various times t (in seconds)
as follows:
t 0 20 40 60 80 100 120
19.3 14.2 10.4 7.6 5.6 4.1 3.0
Plot these results on semi-logarithmic paper and test whether or not
e follows the compound interest law. If so, determine the value of
So and 6 in the equation 6 = SolO"". Note that the last point given
by the table, namely t = 120, 6 — 3.0, goes into a new square if the
scale AB be called 0—100. If the scale AB be called 0—200 then all
entries can appear on a single sheet of form Af5.
2. Graph the following on semi-logarithmic paper:
n
\n
1
li
2
2h
3
w
1.6
3.0
5.1
8.0
14.0
23.0
Show that the equation connecting n and wis w = lO"'"".
Sttggestion: The scale AB, form Mb, may be called 0 — 5 for the
purpose of graphing n.
3. Graph the following on semi-logarithmic paper, and find the
equation connecting n and w.
n
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
w
2.60
3.41
4.45
5.75
7.56
9.85
13.0
16.6
4. A circular disk is suspended in a horizontal plane by a fine wire
at its center. When at rest the upper end of the wire is turned by
means of a supporting knob through 30° The successive angles of
the torsional swings of the disk from the neutral point are then read
at the end of each swing as follows:
Swing number
1
■ 2
3
4
6
6
7
26°.4
23°.2
20°.5
18°.0
15°.9
14°.0
12°.3
Angle
Show that the angle of the successive swings follows the compound
interest law and find in at least two different ways the equation con-
necting the number of the swing and the angle. Show also by the
slide rule that the compound interest law holds. [tt> - SOlO"-"""]
166. The Power Function Compared with the Exponential^
Function. It has been emphasized in this book that the funds-
§156] LOGARITHMIC AND EXPONENTIAL FIJNCTIONS 287
mental laws of natural science are three in number, namely: (1)
the parabolic law, expressed by the power function y = ax"
where n may be either positive or negative; (2) the harmonic or
periodic law, y = asin nx, which is fundamental to all periodically
occurring phenomena; and (3) the compound interest law dis-
cussed in this chapter. While there are other important laws and
functions in mathematics, they are secondary to those expressed
by these fundamental functions. The second of the functions
above named wUl be more fully discussed in the chapter on waves.
The discussion of the compound interest law should not be closed
without a careful comparison of power functions and exponential
functions.
The characteristic property of the power function
y = ax" (1)
is that as x changes hy a constant factor, y changes by a constant
factor also. Let
y = ax" = f(x). (2)
Let X change by a constant factor m, so that the new value of x
is mx. Call y' the new value of y. Then
y' = a{mx)" = f{mx). (3)
That is,
y' a(mx)" ,.,
— = -^^ — '- = m", (4)
y ax"
which shows that the ratio of the two y's is independent of the value
of X used, or is constant for constant values of m.
Another statement of the law of the power function is: As .t
increases in geometrical progression, y, or the power function, in-
creases in geometrical progression also.
r
Let m be nearly 1, say 1 + t^, where r is the percent change in x
and is small, then we have
y' K^' + m) '^K^)'
- = ' ,, ;""' = ^ ^^^^^ = (1 + r)" =F 1 + nr (5)
y fix) ax"
by the approximation formula for the binomial theorem (§113).
288 ELEMENTARY MATHEMATICAL ANALYSIS [§166
Hence,
d' + m)-^^^^
y fix) 100
(6)
f^)=nr. (7)
100
y
The left-hand member is the percent change in y or infix). The
number r is the percent change in the variable a;. Therefore
(7) states that for small changes of the variable the percent of
change in the function is n times the percent of change in the variable.
Let the exponential function be represented by
y = ae'' = Fix). (8)
As already noted in the preceding sections, increasing x by a con-
stant term increases y, or the function, by a constani factor. Thus
y' F{x + h) aeoi'^") '
y Fix) ~ oe»' ~ ' ^ '
which is independent of the value of x, or is constant for constant h.
The expression e'* is the factor by which y or, the function, is in-
creased when X is increased by the term, or increment, h. See
§147.
In other words, as x increases in arithmetical progression, y,
or the exponential function, increases in geometrical progression.
The percent of change is
[Fix + /i) - Fix)-
100
^ = 100 [e'* - 1], (10)
Fix)
which is constant for constant increments h added to the variable x.
If X change by a constant percent from a; to a; ( 1 + t?^) , it will
be found that the percent change in the function is not constant,
but is variable.
The above properties enable one to determine whether measure-
ments taken in the laboratory can be expressed by functions of
either of the types discussed; if the numerical data satisfy the
test that if the argument change by a constant factor the function
also changes by a constant factor, then the relation may be repre-
sented by a power function. If, however, it is found that a change
§157] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 289
of the argument by a constant increment changes the function
by a constant factor, then the relation can be expressed by an
equation of the exponential type.
We have already shown how to determine the constants of the
exponential equation by graphing the data upon semi-logarithmic
paper. In case the equation representing the function is of the
form
y = ae^'' + c, (11)
then the curve is not a straight line upon semi-logarithmic paper.
If tabulated observations satisfy the condition that the function
less (or plus) a certain constant increases by a constant factor as
the argument increases by a constant term, then the equation of
the type (11) represents the function and the other constants can
readily be determined.
The determination of the equations of curves of the parabolic
and hyperbolic type is best made by plotting the observed data
upon logarithmic coordinate paper as explained in the next section.
157. Logarithmic Coordinate Paper. If coordinate paper be
prepared on which the uniform x and y scales are both replaced
by non-uniform scales divided proportionately to log x and log y,
respectively, then it is possible to show that any curve of the para-
bolic or hyperbolic type when drawn upon such coordinate paper will
be a straight line. This kind of squared paper is called logarithmic
paper, and is illustrated in Fig. 116.
To find the equation of a line PQ on such paper, we imagine, as
in the case of semi-logarithmic paper, that aU rulings are erased
and replaced by continuations of the uniform scales ON and MN,
on which the length ON or MN is taken as unity. Denoting, as
before, distances referred to these uniform scales by capital letters,
we may write as the general equation of a straight line
Y = mX + B. (1)
In the case of the line PQ, m = 0.505, B = 0.219, and hence
Y = 0.505X + 0.219.
But, Y = log y, X = log X, where y and x represent distances
19
290 ELEMENTARY MATHEMATICAL ANALYSIS [§157
measured on the scale LM and LO respectively, and 0.219 =
log 1.65. Hence
log y = 0.505 log X + log 1.65
or
log y — log 1.65 = 0.505 log .r.
0 ar 1
10 AT
1
10
u
Q
^
^
^
^
y
y
^
^
^
y
5
^
^
^
4
p
N
:
2
L
1
»0
1 2 3 4567S9 10
Single LogarLtlimic, Scale of Oommon LoearUlims In Margins
Fig. 116. — Logarithmic coordinate paper, form itf4. The finer
rulings of form Af 4 are not reproduced.
This may be written in the form
logj^ = log a;"
whence
or
y — T.0.505
1.65 ~ "" •
y = 1.65a;<'-=''^
(2)
I
§157] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 291
In general, if £ = log 6, we may write the equation (1) in the
form
y — bx'" (3)
If the straight line on logarithmic paper passes through the
point (1, 1) its Cartesian equation is
Y = mX, (4)
or, referred to the logarithmic scales,
log y = m log X — log a;"",
or
y = X". (5)
If the straight line on logarithmic paper passes through the point
(a, 6) with slope m, its equation referred to the logarithmic scales
is
(6)
I = [;]■
On logarithmic paper, form Mi, the numbers printed in the
lower and in the left margin refer to the non-uniform scale in the
body of the paper. By calling the left-hand lower corner the
point (1,10), (10, 10), (10, 1), (10, 100), (1,100) or (100, 100), . . . ,
instead of (1,1), these numbers may be changed to 10, 20, 30,
. , or to 100, 200, 300, . . . , etc.
If the range of any variable is to extend beyond any of the single
decimal intervals, 1—10, 10—100, 100—1000, . . . , the "multiple
paper," form MQ, may be used, or several straight lines may be drawn
across form JW4 corresponding to the value of the function in each
decimal interval, 1 — 10, 10 — 100, . . ., so that as many straight
lines will be required to represent the function on the first sheet as
there are intervals of the decimal scale to be represented. However,
if the exponent m in i/ = bx" be a rational number, say n/r, then the
lines required for all decimal intervals will reduce to r different straight
lines.
One of the most important uses of logarithmic paper is the de-
termination of the equation of a curve satisfied by laboratory
data. If such data, when plotted on logarithmic paper, give
a straight line, an equation of the form (6) above satisfies the
observations and the equation is readily found. The exponent
m is determined by measuring the slope of the line with an ordinary
292 ELEMENTARY MATHEMATIQAL ANALYSIS [§157
uniform scale. The equation of the line is best found by noting
the coordinates of any one point (o, 6) and substituting these
and the slope m in equation (6).
Illustration 1. Construct the semi-cubical parabola y= 2x1
on logarithmic paper.
The result is a straight line of slope -| cutting the line LM, Fig. 116,
at the point marked 2.
100? ^ ■» 9 ~ I
90
80
70
60
EO
40
30
20
£__________.£/ E
/ D F
2 3 4 5 6 78910 20 30 40 5060
K ? M
Fig. 117. — Multiple logarithmic paper.
100
Illustbation 2. Find an empirical equation connecting the x and
y of the accompanying table.
X
y
X
y
5
1.0
20
16.4
7
2.0
30
37.0
9
3.3
40
65.0
15
9.2
50
100.0
These points are shown plotted on the multiple logarithmic paper
in Fig. 117, as the line PQ. The slope of this line is found to be 2.
§157] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 293
Substituting for a and 6 in (fi) the coSrdinates of any of the points
on the line, for example (5, 1), we get
y. = M '
1 \6/
or 2/ = 25^-
Exercises
Draw the following on single or multiple logarithmic paper, forms
M4or Af6:
1. y = x,y =2x,y =3x,y = ix. i. y = x^^, y = x^^,
■.^
2. y = X, y = x^, y = x\ y ^ x*. 5. y = 2x', y = Ja;', A = itrK
Z. y = 1/x, y = l/x\ y = 1/xK
6. Find the empirical equation connecting x and y of the following
table.
X
y
X
y
1.6
3.05
6.5
6.40
2.5
3.92
7.5
6.85
3.5
4.65
8.5
7.25
4.5
5.30
9.5
7.70
5.5
5.82
7. Find the empirical equation connecting x and y of the following
table.
X
y
X
y
1.2
2.15
2.0
5.90
1.3
2.50
2.3
7.80
1.5
3.85
2.5
9.30
1.7
4.30
8. Find the empirical equation connecting x and y of the following
table.
X
y
X
y
1.5
10.0
4.5
3.30
2.0
7.5
5.0
2.98
2.5
6.0
6.0
2.49
3.0
5.0
7.0
2.12
3.5
4.25
8.0
1.87
4.0
3.73
9.0
1.65
294 ELEMENTARY MATHEMATICAL ANALYSIS [§157
Draw the following on single or multiple logarithmic paper as best
suits the particular example. Carefully, label the scales and indicate
the true numerical value of the division points. Use common sense
values of the variables — ^for example in exercise 16 do not graph for
speed over 30 knots.
9. p = 0.003«^, where p is the pressure in pounds per square foot
on a flat surface exposed to a wind velocity of v miles per hour.
Suggestion: The "common sense" range for v is from w = 10 to
V = 100.
c
' 1 G,
.2 .;
1
,'
F
.5
r
.6
r
.8
1
9
iB
f
9
\
\
/
8
/
/
7
\(
A
6
f V
/
\^
5
/
\
/
\
/
^
<,
/
^
K
3
Sj
-.5
>
\
n
s
N
Iv
. o
E
1
\-.2
C
H
-.1
A
1
i
I
a
<
i
5
r
8 9
lOo
Fig. 118.— Diagram for Exercise 10.
10. Find the equ^,tions in rectangular coordinates of the lines EF
and GH of Fig^ 118.
11. V = c-\/rs for c = 110 and r = 1.
12. / = y/2gh for g = 32.2.
13. C = E/B where E = 110 volts.
14. s = Igt^ where g = 32.2.
16. T = TrVZ/g, where g = 32.2.
16. p/po = (p/po)^*"', where po = 0.075, the weight of 1 cubic
foot of air in pounds at 70° F. and at pressure po of 14.7 pounds per
square inch.
§158] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 295
17. H = — p— , for D = 5000, 10,000, 15,000, 20,000, where C =
225, D is displacement in tons and /S is speed in knots.
18. H = -go", for N = 100, 200, 300, 400, 500, 600, 700, 800, 900,
1000. d is the diameter of cold rolled shafting in inches. The line
should be graphed for values of d between d = 1 and d = 10.
19. F = O.OOOSilWBN', where N is revolutions per minute, R is
radius in feet, W is weight in pounds, and F is centrifugal force in
pounds.
'•
/'^
; ;.::;:p
'
y
1
0.9
0.8
=
E
—
-7
2
'','
;';■: ::*
......
;: : : :
^
T
0.0
^
y
;2
^^-
;':';; ;
12
.'
1 Sj .
.'
. ^^
c' t' ■' %
0.R
i
■=;-::^
= !■■
^
;=-:
="^:^ is
.::.
E
~ 7.^
"2*^
- Ji. : . J^ ^
—
^^>
;^:: : :
-■^.
, •
^
<-ii
>
<
<'
0.1
a09
:
:
1
y
;;;:: ;:;
',-;
1
1
i::: : _:
=
0.07
■p
2
z r =
-^ ,. ,. z
-
i'! ::!I -
0.04
X-''
2^: :. I_
«
a2 0.3 a4
£=I*iigthofCreBtinJBet
^=Heaa on Creat in Feet
g =BlB0liai^e In Second Feet
ae 0.8 1.0 3 3 4 6 6 7 8 910
* 0.7 0.9
Discharge over Trapezoidal Wiex
Fig. 119. — A weir formula graphed on multiple logarithmic pa per.
20. g = 3.37M^ for L = 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9,
1.0. See Fig. 119.
O.SSF'*^
21. H = — jj-yjj — , where V is the velocity of water in feet per
second under the head of H feet per 10,000 feet in clean cast-iron pipe
of diameter d feet. See Fig. 120.
158. Slims of Exponential Functions. Functions consisting of
the sum of two different exponential functions are of frequent
occurrence in the application of mathematics, especially in elec-
trical science. Types of fundamental importance are e" + e""
and e» — e-" which are so important that the forms (e" + e~")/2
and (e" — e~")/2 have been given special names and tables of
296 ELEMENTARY MATHEMATICAL ANALYSIS [§158
their values have been computed and printed. The first of
these is called the hyperbolic cosine of u and the second is called
the hyperbolic sine of u ; they are written in the following notation :
cosh w = (6« + e~")/2, sinh li = (e« — e~")/2.
Triction Head in Feei per IDOO Ft. of Pipe
Not«:
For opoD ooadults, multlplj IlydraUllo Radlut bj 4 to
get Equivalent SUmetsT. Diagram givu noarly Boms
reaults aa KuttWB Fonnulm Kith n=.011.
Fur old or foul pipes multlplj required head bj 1.4&
f>00 ^ ^^^ o' divide diagram veloolt; b; I>20 to 1,28 for
V= 2 to & feet per Beoond . j
g For Bubb pipes ffsO-fiO^lia
Diagram, of Flow In Clean Oaat Iron or Wrought Iron Pipes
Baaed on the Formula, H, in Feet per 1000 Feet = 0,38 K"^-
FiG. 120. — A compHcated example of the use of multiple logarithmic paper, Form
MQ. From Transactions Am. Soc. C. E., Vol. LI.
If X = a cosh u and y= a sinh u, then squaring and subtracting
x^ — y^ = a2(cosh'' u — sinh^ u)
=..p
+ 2 + e-2"
2 + e--'
4 4
Therefore the hyperbolic functions
x= a cosh u, and y = a sinh u
'1
§158] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 297
appear in the parametric equations of a rectangular hyperbola
just as the circular functions
X = a cos B, and y = asind
appear in the parametric equations of the circle
a;2 -)- j/2 = o^
4
1
L
3.6
yi
—
\
3
\
/
\
2.5
w
'
I
2
f
*^
1,5
^
^1
^
V'
V
• y
e"
N
\5
l/
V
-yie-
■
X
^
1 — '
^
-3.5
-3
-2.
-2
-1.
-1
H
0
6
1
1.5
2
2.S
3
3.5
/
/
l.S
1
-2
/
2.5
/
-3
/
3.5
/
-4
4.5
-5
Fig. 121. — The curves of the hyperboUc sine and cosine.
The graphs of y = a cosh x and y = a sinh x were called for in
exercises 1, 2, §146. They are shown in Fig. 121. The first
of these curves is formed when a chain is suspended between two
points of support; it is called the catenary. These two curves
are best drawn by averaging the ordinates oi y = e' and y —e~',
and the ordinates oi y = e'' and y = — e"'.
Curves whose equations are of the form
y = ae""* + be"'
take on quite a variety of forms for various values of the constants.
A good idea of certain important types can be had by a comparison
of the curves of Figs. 122 and 123.
298 ELEMENTARY MATHEMATICAL ANALYSIS [§158
1.&
(1)
I
.75
\w
(1) 2^=e-*+o,Be-"'*
(2) 3/=e"" .JO,
(4) y=e:Z-ojseJ"'
(6)2/=e"-i.Be *°*
bA(s)
w
5
P
\
-
25
m
\
N^
^^
— ^
1 1.5
Pig. 123.
2.6
§159] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 299
The student should arrange in tabular form the necessary
numerical work for the construction of the curves of Figs. 122
and 123.
If the coefficient of the second exponent be increased in absolute
value, the points of intersection with the F-axis remain the same,
but the region of close approach of the curves to each other is
moved along the curve y = e-' to a point much nearer the Y-axis
as can be seen by comparing Fig. 123 with Fig. 122.
159. *Damped Vibrations. If a body vibrates in a medium like
a gas or liquid, the amplitude of the swings are found to get smaller
and smaller, or the motion slowly (or rapidly in some cases) dies out.
In the case of a pendulum vibrating in oil, the rate of decay of the
amplitude of the swings is rapid, but the ordinary rate of the decay of
such vibrations in air is quite slow. The ratio between the lengths
of the successive amplitudes of vibration is called the damping factor
or the modulus of decay.
The same fact is noted in case the vibrations are the torsional
vibrations of a body suspended by a fine wire or thread. Thus a
viscometer, an instrument used for determining the viscosity of
lubricating oils, provides means for determining the rate of the decay
of the torsional vibration of a disk, or of a circular cylinder suspended
in the oil by a fine wire. The "amplitude of swing" is in this case the
angle through which the disk or cylinder turns, measured from its
neutral position to the end of each swing.
In all such cases it is found that the logarithms of the successive
amplitudes of the swings differ by a certain constant amount or, as
it is said, the logarithmic decrement is constant. Therefore the
amplitudes must satisfy an equation of the form
A = ae~^
where A is amplitude and / is time. The actual motion is given by an
equation of the form
y = ae~^ sin ct,
A study of oscillations of this type will be taken up more fully in
the calculus. For the present it will suffice to graph a few examples.
Let the expression be
y = g-f/B sin t. (1)
A table of values of t and y must first be derived. There are three
ways of proceeding; (1) Assign successive values to t urespective of
300 ELEMENTARY MATHEMATICAL ANALYSIS [§169
the period of the eine (see Table V and Fig. 124). (2) Select for the
values of t those values that give aliquot parts of the period 2t of the
sine (see Table VI and Kg. 125). (3) Draw the sinusoid y = sin t
carefuUy to scale by the method of §56; then draw upon the same
V
/
'^
s
.
U>b
/
\
\
v'^
A
r
\
^^1,0 1
1 12 4"
t
U
>
V
[^
t
■ J
'i
-'
■^13
0.B
\
/
1.6
V
\
-1.6
Fig. 124. — The curve y = e"*/' sin t.
\
s*.
/
N
>N
t"
/
\
<1
r^
\
TT
ITT
/^
'n
lir
4T
J
(
V
i 1
[/
2 1
4 1
6J
M
U2
^
4 2
B 2
\
>^
^
^
rr"'
/
Fig. 125. — The curve y = e"*/' sin t.
coordinate axes, using the same units of measure adopted for the sinus-
oid, the exponential curve y = e~'/^; finally multiply together, on the
slide rule, corresponding ordinates taken from the two curves, and
locate the points thus determined.
§159] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 301
The first method involves very much more work than the second
for two principal reasons: First, tables of the logarithms of the
trigonometric functions with the radian and the decimal divisions
of the radian as argument are not available; for this reason 57.3°
must be multiplied by the value of t in each case so that an ordinary
trigonometric table may be used; second, each of the values written
in column (3) of the table must be separately determined, while if
the periodic character of the sine be taken advantage of, the numerical
values would be the same in each quadrant.
TABLE V
Table of the function y
= e "^ sin t
1
2
3
4
5
t in radians
log e-'/» =
- (0.0869)i
log sin ( or log
sin 57.34 if ( is
in degrees
logy
V
0.0
-0.0000
+ 0.000
0.5
- 0.0434
9.6807
9.6372
+ 0.434
1.0
- 0.0869
9.9250
9.8381
+ 0.689
1.5
- 0.1303
9.9989 '
9.8686
+ 0.739
2.0
-0.1737
9.9587
9.7850
+ 0.610
2.5
- 0.2172
9.7771
9.5599
+ 0.363
3.0
- 0.2606
9.1498
8.8892
+ 0.077
3.5
- 0.3040
9.5450
9.2410
-0.174
4.0
-0.3474
9.8790
9.5312
- 0.340
4.5
-0.3909
9.9901
9.5992
- 0.397
5.0
- 0.4343
9.9818
9.5475
- 0.353
5.5
- 0.4777
9.8485
9.3708
- 0.235
6.0
- 0.5212
9.4464
8.9252
- 0.084
6.5
- 0.5646
9.3322
8.7679
+ 0.059
7.0
- 0.6080
9.8175
9.2095
+ 0.162
7.5
- 0.6515
9.9722
9.3207
+ 0.209
8.0
- 0.6949
9.9954
9.3005
+ 0.200
8.5
- 0.7383
9.9022
9.1634
+ 0.146
9.0
- 0.7817
9.6149
8.8332
+ 0.068
9.5
- 0.8252
8.8760
8.0508
-0.011
10.0
- 0.8686
9.7356
8.8670
-0.074
10.5
- 0.9120
9.9443
9.0323
- 0. 108
11.0
- 0.9555
9.9999
9.0444
-0.111
11.5
- 0.9989
9.9422
8.9433
- 0.088
12.0
- 1.0424
9.7296
8.6872
-0.049
The second method, because of the use of aliquot divisions of the
302 ELEMENTARY MATHEMATICAL ANALYSIS [§159
period of tl^e Bine, such as ir/6 or ir/12 or r/l& or 5r/20, etc., possesses
the advantage that the values used in column (3) need be found for one
quadrant only and the values required in column (2) are quite as
readily found on the slide rule as in the first method.
TABLE VI
Table of the function y
= e""/* sin t
1
2
3
4
5
n = t in
units of
ir/6 radians
log e- »"■/'" =
- (0.0455)»
log, ain nw/6
log!/
y
0
1
-0.0000
- 0.0455
0.000
+ 0.450
9.6990
9.6535
2
-0.0910
9.9375
9.8465
+ 0.702
3
- 0.1364
0.0000
9.8636
+ 0.731
4
- 0.1819
9.9375
9.7556
+ 0.570
5
- 0.2274
9.6990
9.4716
+ 0.296
6
-0.2729
+ 0.000
7
- 0.3184
9. ,6990
9.3806
- 0.240
8
- 0.3638
9.*9375
9.5737
- 0.375 _
9
- 0.4093
0.0000
9.5907
-0.390
10
-0.4548
9.9375
9.4827
- 0.304
11
- 0.5003
9.6990
9.1987
-0.158
12
13
- 0.5458
-0.5912
0.000
+ 0.128
9.6990
9.1078
14
- 0.6367
9.9375
9.3008
+ 0.200
15
- 0.6822
0.0000
9.3178
+ 0.208
16
-0.7277
9.9375
9.2098
+ 0.162
17
-0.7732
9.6990
8.9258
+ 0.084
18
19
- 0.8186
-0.8641
0.000
- 0.068
9.6990
8.8349
20
- 0.9016
9.9375
9.0279
-0.107
21
- 0.9551
0.0000
9.0449
- 0.111
22
- 1.0006
9.9375
8.9369
-0.087
23
- 1.0460
9.6990
8.6530
-0.045
24
- 1.0915
0.000
The third method is perhaps more desirable than either of the
others if greater than two figures accuracy is not required. The
curve can readily be drawn with the scale units the same in both
dimensions, as is sometimes highly desirable in scientific applications.
In Figs, 124 a^d 125 a, larger unjt ha? be^n used on the vertical
§159] LOGARITHMIC AND EXPONENTIAL FUNCTIONS 303
scale than on the horizontal scale. In Fig. 125 the horizontal unit
is incommensurable with the vertical unit. To draw the curve to a
true scale in both dimensions it is preferable to lay off the coSrdinates
on plain drawing paper and not on ordinary squared paper. Rec-
tangular coordinate paper is not adapted to the proper construction
and discussion of the sinusoid, or of curves, like the present one, that
are derived therefrom.
Curves whose equations are of the form y = je"*/' sin tor y =
Se"'.'' sin t, etc., are readily constructed, since the constants i, 3, etc.,
merely multiply the ordinates of (1) by \, 3, etc., as the case may be.
Likewise the curve y = e~'* sin ex is readily drawn since sin ex can be
made from sin x by multiplying all abscissas of sin x by 1/c.
CHAPTER X
TRIGONOMETRIC EQUATIONS AND THE SOLUTION
OF TRIANGLES
A. FURTHER TRIGONOMETRIC IDENTITIES
160. The circle p = a cos d + h sin 0. In §74 an analytical
proof was given of the fact that p = a cos fl + 6 sin 5 is the polar
equation of a circle passing through the pole 0 and having its
center at the point (\a, 56). The demonstration there given
should now be reviewed.
Geometrical Explanation. The following geometrical dis-
cussion should give the
student a better under-
standing of the important
theorem of §74.
We know (§66) that pi
= a cos B is the polar equa-
tion of a circle of diameter
a, the diameter coinciding
in direction with the polar
axis OX; for example, the
circle OA, Fig. 126. Like-
wise, p2 = 6 sin 0 is a circle
whose diameter is of length
6 and makes an angle of
-|-90° with the polar axis
OX, for example, the circle
OB, Fig. 126. Also, p =
c cos {0 — 6) is a circle whose
diameter c has the direction angle 5. See Theorem XIV on Loci,
§70. We shall show that if the radius vectors corresponding to
any value old in the equations pi = a cos d and pa = 6 sin 6 be added
together to.make a new radius vector p, then, for all values of B,
304
Pig. 126. — Combination of the cir-
cles p = a cos 6 and p = 6 sin 9 into a
single circle p = a cos B + 6 sin e.
§160] TRIGONOMETRIC EQUATIONS 305
the extremity of p lies on a circle (the circle OC, Fig. 126) of di-
ameter Vo^ + h'^. In other words we shall show geometrically that
p = a cos & + 6 sin fl (1)
is the equation of a circle.
In Fig. 126, pi = a cos Q wiU be called the a-cirde OA; p^ =
6 sin 6 will be called the b-circle OB. For any value of the angle
6 draw radius vectors OM, ON, meeting the a- and 6-circles respec-
tively at the points M and N. If P be the point of intersec-
tion of MN produced with the circle whose diameter is the diagonal
OC of the rectangle described on OA and OB, we shaU show that
OM + ON = OP, no matter in what direction OP be drawn.
Let the circle last mentioned be drawn, and project BC on OP.
Since ONB and OPC are right angles, NP is the projection of
BC {= a) upon OP. But OM also is the projection of a (= OA)
upon OP. Hence NP = OM because the projections of equal
parallel lines on the same line are equal. Therefore, for all values
of d, NP = pi and OP = ON + NP = pi + pi, which is the fact
that was to be proved.
Designating the angle AOC by 5, the equation of the circle OC is
by §70.
P = Va^ 4- h'^ cos {6 - S) (2)
The value of 5 is known, for its tangent is -• It should be observed
that there is no restriction on the value oi 6. As the point P
moves on the circle OC, the circumference is twice described as d
varies from 0° to 360°, but the diagram for other positions of the
point P is in no case essentially different from Fig. 126.
The above reasoning and the diagram involve the restriction
_ that both o and 6 are positive numbers. While it is possible to
supplement the reasoning to cover the cases in which this restric-
tion is removed, it is unnecessary as the analytical proof of §74
is applicable for all values of a and b.
The equation of the circle OC in any position, that is, for
any value of a and 6, positive or negative, may also be written
in the form
p = Va^ + b^sm{d + i) ' (3)
in which e is the angle BOC in Fig. 126. The equation of the
20
306 ELEMENTARY MATHEMATICAL ANALYSIS [§160
circle OC has therefore been written in three different forms,
namely equations (1), (2) and (3) above.
Illustration 1. From the above we know that the equation
p = 6 cos 9 + 8 sin 9 represents a circle. The diameter of the
circle is Va* + ft" = VSM"^ = lOi so that the equation of the
circle may also be written p = 10 cos (9 — i), where 5 is the angle
whose tangent is - = ^ = 1.33. From a table of tangents S = 53° 8',
so that the equation of the circle may be written p = 10 cos (9 — 5l3 °8').
Illustration 2. Write the equation of the circle p = cos 9 —
-y/S sin 9 in the form p = c cos (9 — S) and in the form p =
c sin (9 + e).
Here a = 1, 6 = — -y/S, c = \/a* + 6« = 2. Hence C must be the
point (1, — \/3) in the second quadrant. Then 5 = angle of second
quadrant whose tangent is ( — -s/S/l), or 120°. Also « = — 30°. Hence
the required equations are p = 2 cos (9 — 120°) and p = 2 sin (9 — 30°).
The result of this section should also be interpreted when the vari-
ables are x and y in rectangular coordinates, and not p and 9 of
polar coordinates. Thus, y = a cos a; is a sinusoid with highest point
or crest at a; =0, 2t, iar, . . . Likewise, y = b sin s is a sinusoid
with crest at a; = y -«-' -~-' . . The above demonstration shows
that the curve
y = a cos X + b sin X
is identical with the sinusoid
y = Va' + 6" cos (a; - hi) = \/a'- + ¥ sin (a; + h)
of amplitude \/a^ -\- b- and with the crest located at a; = hi, or at
s — ^2, where hi is, in radians, the angle whose tangent is -> and hi
is, in radians, the angle whose tangent is r-
Exercises
1. Put the equation p = 2 cos 9 + 2v'3 sin 9 in the form p =
c cos (9 — 8) and find the value of h.
2. Put the equation p = 4 cos 9 + \\/Z sin 9 in the form p =
ccos (9 — 5).
3. Put the equation p = — 4 cos § — 4 sin 9 in the form p -
c sin (9 + e),
§161] TRIGONOMETRIC EQUATIONS 307
4. Put the equation p = 2-\/3 cos 6 + 2 sin fl in the form p =
c cos (9 — 8).
6. Put the equation p = 3 cos 9 + 4 sin 9 in the form p =
csin (9 + e). Put the same equation in the form p = c cos (9 — 'S).
(S is the angle AOC, Fig. 126.)
6. Put the equation p = 5 cos 9 + 12 sin 9 in the form p =
o sin (9 + e); also in the form p = c cos (9 — S).
7. Put the equation (x — 1)^ + (y — ly = 2 in. the form p =
c sin (9 + a) and determine c and a.
8. Put the equation (a; + 1)^ + (2/ — \/3)' = 4 in the form p =
c sin (9 — a) and determine c and a.
9. Put the equation (x + 1)^ + (y + Vs)^ = 4 in the form p =
c sin (9 — a) and determine c and a.
10. Find the maximum value of cos 9 — \/3 sin 9, and determine
the value of 9 for which the expression is a maximum.
Suggestion : Call the expression p. The maximum value of p is the
diameter of the circle p = cos 9 — -y/s sin 9. The direction angle of
the diameter is the value of a when the equation is put in the form
p = c cos (9 — a).
11. Find the value of 9 that renders p = f-v/S cos 9 — ^ sin 9 a
maximum and determine the maximum value of p.
12. Find the maximum value of 3 cos t + isint.
161. Addition Fonnulas for the Sine and Cosine. From the
preceding section, equations (1), (2) and (3), we know that the
equation of the circle OC, Fig. 127, may be written in any one of
the forms
p = a cos 6 + 6 sin 0, (1)
p = c sin(,e- e), (2)
p = c cos (e-5). (3)
Hence, for all values oi 9, d, and e,
sin (6 — e) = - cos 5 + - sin 6, (4)
cos (9 - 5) = -cose + - sin 9, (6)
In each of these equations c = -^/a" + h^. The letters a and 6
stand for the co6rdinates of C irrespective of their signs or of
the position of C,
308 ELEMENTARY MATHEMATICAL ANALYSIS [§161
Since (4) and (5) are true for all values of 6, they are true when
» = 0° and when 0 = 90°.
First, Let 0 = 0° in (4).
Then a/c = sin (— e) = — sin e by §60, (6)
-Second, let B = 90° in (4),
Then 6/c = sin (90° - e) = cos « (7)
Substituting (6) and (7) in (4) we have
sin (5 — e) = sin 0 cos e — cos 6 sin e (8)
Fig. 127. — The circle p = c cos (9 — 5) or p = sin c (fl — e) used
in the proof of the addition formulas. Note that e = 90° + «
which is also true for negative angles, namely Si = 90° + ei.
In like manner upon letting 0 = 0 and 0 = 90° in succession in
(5) we have
- = cos (- 5) = cos 5, by §60. (9)
= cos (90 — 5) = sin 5.
Substituting (9) and (10) in (5) we obtain
cos {8 — B) = cos 9 cos 6 + sin 0 sin 5
(10)
(11)
§162] TRIGONOMETRIC EQUATIONS 309
Since these are, true for all values of S and e, put 5 = (-^Si) and
e = ( — ei). Then by §60, these equations become
sin (6 + ei) = sin 0 cos ei + cos 6 sin €i (12)
cos (0 + Si) = cos 6 cos 5i — sin 9 sin 5i (13)
To aid in committing these four important formulas (8), (11),
(12) and (13) to memory, it is best to designate in each case the
angles by a and |3, and write (12) and (13) in the form
sin (a + /3) = sin o: cos /3 + cos a sin /3 (14)
cos (a + /3) = cos a cos |8 — sin a sin /3 (15)
and also write (8) and (11) in the form
sin (a — P) = sia a cos /3 — cos a sin /3 (16)
cos (a — |3) = cos a cos /3 + sin q: sin j3 (17)
The four formulas (14), (15), (16) and (17) must be committed to
memory. They are called the addition fonnulas for the sine and
cosine. The above demonstration shows that the addition for-
mulas are true for all values of a and fi.
By the above formulas it is possible to compute the sine and
cosine of 75° and 15° from the following data:
sin 30° = i sin 45° = iV2
cos 30° = i V3 cos 45° = iV2
Thus
sin 75° = sin (30° + 45°) = sin 30° cos 45° + cos 30° sin 45°
= HV2 + |\/3iV2 = iV^(V'3 + 1)
Likewise
sin 15° = sin (45° - 30°) = i-s/2(\/3 - 1)
162. Addition Formula for the Tangent. Dividing the mem-
bers of (14) §161 by the members of (15) we obtain
, , a\ sin (a -t- /3) sin a cos |8 -h cos a sin |3 ,, ,
tan (a + p) = -, — —37- = 5 ; ; — 5 ^1;
cos (o -t- p) cos acosp — sm asmp
Dividing numerator and denominator of the last fraction by
cos a cos /3
sin a cos /3 cos a cos |8
tan ia + ^) = ^^E^^l_Jo[^^ (2)
cos a cos fi _ sin a sin fi
cos a cos ^ cos acosfi
310 ELEMENTARY MATHEMATICAL ANALYSIS [§163
or
, , .. tan a + tan /3 ,„,
tan (a + B) = — ^ (3)
y I t^' I - tan<»tan|3
Likewise it can be shown from (16) and (17), §161, that
. , „ tana — tan ^ , .
tan (a - |3) = — t— r— ^ (4)
^ ' I + tanatan|3
Equations (3) and (4) are the addition formulas for the tangent.
Exercises
1. Compute cos 75° and cos 15°.
2. Compute tan 75° and tan 15°.
3. Write in simple form the equation of the circle
p = sin 6 + cos B.
4. Put the equation of the circle p = 3 sin 9 + 4 cos 6 in the form
p = c sin (9 + 9i) and find, from the tables or by the slide rule, the
value of ©i.
6. Derive a formula for cot (a + p).
6. Prove cos (s + t) cos {s — t) = cos'' s — sin^ t.
7. Express in the form c cos (a — b) the binomial 3 cos a + 4 sin o.
8. Express in the form c sin (a + 6) the binomial 5 cqs o + 12 sin a.
9. Find the coordinates of the maximum point or crest of the sinus-
oid y = sin X + -\/3 cos x. [First reduce the equation to the form
2/ = c sin (a; + a)].
163. Functions of Composite Angles. The sine, cosine, or
tangent of the angles (90° — d), (90° + 6), (180° - 6), (180° + 6),
(270° — 6), (270° + d) can be expressed in terms of functions
of 6 alone by means of the addition formulas of §§161 and 162.
Thus, write .
sia (a + /?) = sin a cos /3 + cos a sin /3 (1)
cos {a + fi) = cos a cos ;8 — sin a sin j3 (2)
Put a = 180°, and /3 = + 0; then (1) and (2) become, re-
spectively,
sin (180° ± 0) = T sin e (3)
cos (180° ±6) = ~ Gosd (4)
TRIGONOMETRIC EQUATIONS
311
Also in (1) and (2) put a = 90°, and P = ±6, then (1) and (2)
become, respectively,
sin (90° ± 6) = cos 6 (5)
cos (90° ± 6) = + sine ■ (6)
By division of (3) by (4) and of (5) by (6),
tan (180° ±6) = + tan 6», (7)
tan (90° ± 6) = + cot d. (8)
In a similar manner all of the results given in the following table
may be proved to be true.
(-ft./l) Pa
AC ft, A)
(A. ft)
P(h.k)
Pt(.-h,-h)
P,(h.-k)
(-ft.-ft) Pa
P, (k.-h)
A B
Fig. 128. — An angle 9 combined with an even number of right
angles, (A) and wijh an odd number of right angles, (B) .
TABLE VII
Functions of 6 Coupled with an Eeen or with an Odd Number of
Bight Angles
- e
90° -9
90°+ 9
180°- 9
180°+ 9
270°- 9
270°+ 8
sin
— sin 9
cos 9
cos 9
sin 9
— sin 9
— cos 9
— cos B
cos
' cos e
sin 9
— sin 9
— cos 9
— cos 9
— sin B
sin 8
tan
— tan e
cote
— cot e
— tan 9
tan 9
cot 9
- cot 9
AU of the above results can be included in two simple state-
ments. For this purpose it is convenient to separate into different,
312 ELEMENTARY MATHEMATICAL ANALYSIS [§163
classes the composite angles that are made by coupling 0 with
an odd number of right angles, as (90° + fl), (ff - 90°), (270° - 6),
(450° + 6), etc., and those composite angles that are made by
coupling 6 with an even number of right angles, as (180° + 6)
(180° - 6), (360° - 6), (- 6), etc. Note that 0 is an even
number, so that ( — 9) or (0° — 6) falls into the second class of
composite angles. We can then make the following statements :
Theorems on Functions of Composite Angles
Think of the original angle 6 as an angle of the first quadrant:
I. Any function of a composite angle made by coupling B {by
addition or subtraction) with an even number of right angles, is
equal to the same function of the original angle 6, with an algebraic
sign the same as the sign of the function of the composite angle in
its quadrant.
II. Any function of a composite angle made by coupling 0 (by
addition or subtraction) with an odd number of right angles, is eqital
to the co-function of the original angle B, with an algebraic sign
the same as the sign of the function of the composite angle in its
quadrant.
For example, let the original angle be 6, and the composite angle be
(180° + 8). Take any function of (180° + $), say tan (180° + 6), it is
equal to + tan 6, the sign + being the sign of the tangent in the quad-
rant of the composite angle (180° + 8), or third quadrant. Likewise
cot (270° + 6) must equal the negative co-function of the original
angle, or — tan 0, the algebraic sign being the sign of the cotangent in
the quadrant of the composite angle (270° -|- 9), or fourth quadrant.
In the above statements it has been assumed that the angle fl is an
angle of the first quadrant. This is merely for the convenience of
determining signs, for the results stated in itaUcs are true, no matter
in what quadrant 9 may actually, he.
Exercise
Given sin 30° = J, cos 30° = JVS, tan 30° = iVS, cot 30° = \/3,
find the sine, cosine, and tangent of each of the following angles by
means of the above Theorems on Functions of Composite Angles:
(a) 150°; ^b) 210°; (c) 240°; (d) 300°; (e) 330°; (/) 120°; (g) 60°;
(h) -30°.
§164]
TRIGONOMETRIC EQUATIONS
313
164. Angle that a Given Line Makes with Another Line. The
slope m of the straight line y = mx + b is the tangent of the
direction angle, that is, the tangent of the angle that the line makes
with OX. If Li and L^ are any two lines in the plane, the angle
that Li makes with Lj is the positive angle through which L^ must
he rotated about their point of intersection in order that Li may
coincide with Li. Represent the direction angles of two straight
lines
y = miX + bi (1)
y = mix. + hi (2)
by the symbols di and 6^. Then, through the intersection of the
lines pass a line parallel to the OX-axis, as shown in Fig. 129.
Call 0 the angle that the line Li makes with La) that is, the positive
^
'\>
'>
A
0
V
\-
\
\L,.
I Fig. 129. — The angle <l> that a line Li makes with -La. J
angle through which La, considered as the initial line, must be
turned to coincide with the terminal position given by Li. If
9i > Bi, then 4> = Bi- 6^, but if fla > Oi, then 0 = 180° - (Sj - di).
In either case (by equations (7), §163, and (3), §60
tan <i> = tan {di - d^). (3)
That is,
tan di — tan 82 ,,.
(4)
or
tan (b =
tan <j> =
1 + tan 61 tan 62'
nil — nia
(5)
I + niinia
The condition that the given lines (1) and (2) are parallel is
obviously that
mi = ma (6)
Thus, the lines y = 5x + 7 and y = 5x — 11 ar« parallel.
314 ELEMENTARY MATHEMATICAL ANALYSIS [§164
The condition that the given lines (1) and (2) are perpendicular
to each other is that tan 4> shall become infinite; that is, that the
denominator of (5) shall vanish. Hence the condition of perpen-
dicularity is
1 + miW2 = 0,
m: = - ^- (7)
Therefore, in order that two lines may be perpendicular to each
other, the slope of one line must be the negative reciprocal of the slope
of the other line.
Thus the lines y = %x — A and y = — fa; + 2 are per-
pendicular.
Exercises
1. Find the tangent of the angle that the first line makes with
the second line of each set:
[a) y = 2x + Z, y = x + 2,
{h)y = Zx -Z, y = 2x + 1,
(c) y = ix + 5, y = Zx - 1,
id) y = lOx + I, y = Ux - 1,
2. Find the angle that the first line of each pair makes with the
second:
(a) y = X +5, y = - a; -|- 5.
(6) 2/ = Ja; -H 6, y = - 2x.
(c) 2/ = 2a; + 4, y = x + 1.
(d) 2x+Zy = \, Ix =y = \.
(e) 2i 4- 42/ = 3, 3a; 4- 62/ = 7.
(/) 2x +Ay = 3, 6a; - 32/ = 7.
3. Find the angle, in each of the following cases, that the first
line makes with the second: -
(o) 2/= x/Vz +4, 2/ = V3x+ 2.
(6) y = a;/\/3 -|- 1, y = VZx- 4.
(c) y = y/Zx - 6, 2/ = s/Zx- Z.
4. Find the angle that 2i/ — 6a; -1- 7 = 0 makes with y + 2x +
7=0 and also the angle that the second line makes with the first.
§165] TRIGONOMETRIC EQUATIONS 315
166. The Functions of the Double Angle. The addition
formulas for the sine, cosine, and tangent reduce to formulas of
great importance for the special case fi = a.
Thus sin (a + a) = sin a cos a + cos a sin a,
or sin 2a = 2 sin a cos a. (1)
Also cos (a + a) = cos a cos a — sin a sin a
which can be written in the three forms:
cos 2 a = cos^ a — sin^ a, (2)
cos 2 a = 2 cos^ oi —I, (3)
cos 2 o! = I — 2' sin^ a. (4)
Forms (3) and (4) are obtained from (2) by substituting,
respectively, sin^ a = 1 — cos'' a and cos^ a = 1 — sin^ a.
Equations (3) and (4) are frequently useful in the forms :
— , . I + cos 2a; ,g,
(6)
Again
2
•
sin^
a
=
I — cos 2a
2
tan {a + a)
=
tan a + tan a
1 — tana tan a
fan
2 tana
■•""'"" I - tan» a ^^^
166. The Functions of the Half Angle. From (6) and (5) of
§165 we obtain, after replacing a by u/2 and extracting the
square root,
sin (u/2) = ±v'(i — cos u)/2, (1)
cos (u/2) = ± \/(i + cos u) /2 . (2)
Dividing (1) by (2), we obtain
* /- / ^ j_ 1 /i - COS u j^ I - cos u ^ sin u . ,„.
tan (u/2) = ± V — ; = ± ^^ = ± — i (3)
^ ' ' » I + cos u sm u I + cos u
Formulas (1), (2) and (3) have many important applications in
mathematics. As a simple example, note that the functions of 15°
316 ELEMENTARY MATHEMATICAL ANALYSIS [§167
may be computed when the functions of 30° are known. Thus
cos 30° = (1/2) VS
therefore sin 15° = \/(l - cos 30°)/2 = V'l/2 - (1/4)^3-
Also cos 15° = Vl/2 + (1/4)^3.
Likewise by (3)
tan 15° = L^^#^ = 2 - V3.
Exercises
1. Compute sin 60° from the sine and cosine of 30°.
2. Compute sine, cosine, and tangent of 221°.
3. If sin X = 2/5, find the numerical value of sin 2x, cos 2x, and
tan 2x.
4. Show by expanding sin (x + 2x) that sin 3a; = 3 sin a; — 4 sin 'x.
, _, . „ 3 tan x — tan' x
0. Prove tan 3a; = — 5 5-: — ;
1—3 tan* X
6. Show that sin 29/sin e — cos 29/cos 9 = sec 8.
7. Show that
Ism 2 + cos^) = 1 + sin e.
8. Show that cos 29(1 + tan 29 tan B) = 1.
9. If sin A = 3/5, calculate sin {A/2).
10. Prove that tan (7r/4 + 9) = ^ _ ^ g-
11. Prove that tan (ir/4 - 9)' = (1 - tan 9)/(l + tan 9).
12. Show that sec 9 + tan 9 = ^^— •
cos 9
. » p., . , . 1 + 2 sin a cos a cos a + sin a
13. Show that -■ x—- ;-= — = ■ — -.
cos* a — sm* a cos a — sm o
14. Show that sec 9 + tan 9 = tan
[i+g
16. Show that — ^—j — j :r-^ — tan A tan B.
cot A + cot B
16. Prove that cos (s + t) cos {s — t) + sin (s + t) sin (s — i) =
cos 2t.
167. Sums and Differences of Sines and of Cosines Expressed
as Products. The following formulas, which permit the substi-
tution of a product for a sum of two sines or of two cosines, are
§167] TRIGONOMETRIC EQUATIONS 317
important in many transformations in mathematics, especially in
the calculus. They are immediately derivable from the addition
formulas. Thus, by the addition formulas (14) and (16), §161, we
obtain
sin (a + 6) + sin (a — 6) = 2 sin o cos 6.
Likewise by subtraction of the same formulas
sin (a + 6) — sin (a — b) = 2 cos a sin b.
By the addition and subtraction, respectively, of the addition
formulas for the cosine there results
cos (a + 6) + cos (a — b) = 2 cos a cos 6.
cos (a + 6) — cos {a — b) = — 2 sin a sin 6.
These formulas can be written
sin o cos b — 5 [sin (o + 6) + sin (a — 6)], (1)
cos a sin 6 = | [sin (a + 6) — sin (o — b)], (2)
cos a cos 6 = 2 [cos (a + b) + cos (a — 6)], (3)
sin a sin 6 = — 5 [cos (a + 6) — cos (a — 6)]. (4)
Represent (a + 6) by a and (a — b) by /?.
Then o = (a + /3) /2 and b = (a - j8)/2
Hence the above formulas become
sin a + sin j3 = 2 sm cos -' (5)
• n a + 0 . a — 0 ,„,
sm a — sin fl = 2 cos sm > (6)
22
cos a + cos /3 = 2 cos cos > (7)
2 2
cos a — cos /3 = — 2 sm sm ^- (8)
2 2
The principal use of these formulas is in certain transformations
in the Calculus. A minor use is in adapting certain formulas to
logarithmic work by replacing sums and differences by products.
These formulas should not be committed to memory. They
can be derived in a moment when needed by recalling their
318 ELEMENTARY MATHEMATICAL ANALYSIS [§168
connection with the addition formulas. Formula (2) is really-
contained in formula (1). For by (1)
cos a sin 6 = sin 6 cos a
= 5 [sin(& + a) + sin (6 — a)]
= 5 [sin (a + 6) — sin (a — 6)],
since sin(— B) = — sin &
Exercises
Express as the sum or difference of sines or cosines:
1. sin 5x cos 2x. 6. sin 3x sin 7x.
2. cos 3a; sin 7x. 7. cos 3a; cos 8x.
3. cos 4a; cos x. 8. cos 5a; sin 2x.
4. sin 5x sin 2x. 9. sin 3a; cos lOx.
6. sin 3x cos 5x. 10. cos 2x cos 6x.
168.* Graph of y = sin 2x, y = sin nx, etc. Since the substi-
tution of nx for X in any equation multiplies the abscissas of the
curve by 1/n, or («>!) shortens, or contracts, the abscissas of all
points of the curve in the uniform ratio n : 1, the curve y = sin 2x
must have twice as many crests, nodes, and troughs in a given
interval of x as the sinusoid y = sin x. The curve y = sin 2x is
therefore readily drawn from Fig. 59 as follows: Divide the axis
OX into twice as many equal intervals as shown in Fig. 59 and
draw vertical lines through the points of division. Then in the
new diagram there are twice as many small rectangles as in the
original. Starting at 0 and sketching the diagonals (curved to
iit the alignment of the points) of successive cornering rectangles,
the curve y = sin 2x is constructed. It is, of course, the ortho-
graphic projection of J/ = sin x upon a plane passing through
the F-axis and making an angle of 60° (the angle whose cosine is
1/2) with the x2/-plane. The curve y = cos 2x is sunilarly con-
structed. In each of these cases we see that the period of the
function is t and not 2ir.
169.* Graph of p = sia20,p — cos 20, etc. The curve p = cos 6
is the circle of diameter unity coinciding in direction with the axis
OX. We have already emphasized that as d varies from 0° to
360° the circk is twice drawn, so that the curve consists of two
§170]
TRIGONOMETRIC EQUATIONS
319
superimposed circular loops. Now p = cos 2d wiU be found to
consist of four loops, somewhat analogous to the leaves of a four-
leafed clover, but each loop is described but once as 6 varies from
0° to 360°. The curve p = cos 36 is a three-looped curve, but each
loop is twice drawn as S varies from 0° to 360°. Also p = cos 116
has eleven loops, each twice drawn, while p = cos 126 has twenty-
four loops, each one described but once, as 6 varies from 0° to 360°.
The curves p = cos 2 6, p = sin 39, p = sin 6/2 should be drawn
by the student upon polar coordinate paper.
170.* Graph of y = sln^x, and y = cos^x. The graphs y = sin' x
and y = cos'' x have important applications in science. The following
E
Y
A
s
P^-^?
■'^.^-L
—
—
—
—
—
ci._4V.
\
/
N
rr-l M
N^
/
N
ff|--| -
/
s
/ "^ \ \3^
1 1 /
S^
/
"— ^^"^-^ \
/
~~—-^^^~^
<.
f^
^"^^^r
0
y
/
/
H
Fig. 130. — The graph oi y = cos^ x.
graphical method offers an easy way of constructing the curves and it
illustrates a number of important properties of the functions involved.
We shall first construct the curve y = cos' x. At the left of a sheet of
8 J X 11-inch paper draw a circle of radius 36/57r ( = 2.30) inches, (OA,
Fig. 130). Lay off the angles 9 from OA, as initial line, correspond-
ing to equal intervals (10°) of the quadrant APE as shown in the
figure. Let the point P mark any one of these equal intervals.
Then dropping the perpendicular AB from A upon OP, the dis-
tance OB is the cosine of 6, if OA be called unity. Dropping a
perpendicular from B upon OA, the distance OC is cut off, which is
equal to OB' or cos' e, since in the right triangle OB A, OB' — OCOA
= OC-1. Making similar constructions for various values of the angle
e, say for every 10° interval of the arc APE, the hne OA is divided at
a number of points proportionally to cos' e. Draw horizontal lines
320 ELEMENTARY MATHEMATICAL ANALYSIS [§171
through each point of division of OA. Next divide the axis OX into
intervals equal to the intervals of 8 laid off on the arc APE. Since the
radius of the circle OA was taken to be (SB/Sir) inches, an interval of
10° corresponds to an arc of length 2/5 inch, which therefore Inust
be the length of the equal intervals laid off on OX. Through each of
the points of division of OX draw vertical lines, thus dividing the
plane into a large number of small rectangles. Starting at A and
sketching the diagonals of successive cornering rectangles, the locus
ARS oi y = cos' x is constructed.
From Fig. 130, it is seen that B always lies at the vertex of a right-
angled triangle of hypotenuse OA. Thus as P describes the circle of
radius OA, B describes a circle of radius OA/2. Therefore the curve
ABS is related to the small circle ABO in the same manner that
the curve of Fig. 59 is related to its circle; consequently the curve
ARS of Fig: 130 is a sinusoid tangent to the X-axis. Thus the graph
y = cos' a; is a cosine curve of amplitude 1/2 and wave length or period
IT, lying above the X-axis and tangent to it.
B. PLANE TRIANGLES: CONDITIONAL EQUATIONS
171. Law of Sines. The first of the conditional equations per-
taining to the oblique triangle is a proportion connecting the sines
Fig. 131. — Derivation of the law of sines and the law of cosines.
of the three angles of the triangle with the lengths of the respec-
tive sides lying opposite. Call the angles of the triangle A, B, C,
and indicate the opposite sides by the small letters a, b, c, respec-
tively. From the vertex of any angle, drop a perpendicular p
upon the opposite side, meeting the latter (produced if necessary)
§172] TRIGONOMETRIC EQUATIONS 321
at D. Then, by the properties of right triangles, we have, in
either Fig. 131 (1) or 131 (2),
p = c sin DAB. (1)
From A BDC,
p = a sin C. (2)
But,
sin DAB = sin A, Fig. 131 (1)
= sin (180° - A), Fig. 131 (2)
= sin A.
Therefore p = c sin A = » sin C, (3)
or a/sin A = c/sin C. (4)
In like manner, by dropping a perpendicular from A upon a, we
can prove
b/sin B = c/sin C. (5)
Therefore a/sin A = b/sin B = c/sin C (6)
Stated in words, the formula says: In any oblique triangle the
sides are proportional to the sines of the opposite angles.
Geometrically: Calling each of the ratios in (6) 2B, it is seen
from Fig. 131 (2) that R is the radius of the circumscribed circle
since c/ sin C = 2R can be deduced from the triangle BAE. Similar
construction can be made for the angle B or A.
172. Law of Cosines. From plane geometry we have the theo-
rem: The square of any side opposite an acute angle of an oblique
triangle is equal to the sum of the squares of the other two sides di-
minished by twice the product of one of those sides by the projection
of the other side on it. Thus, in Fig. 131 (1),
o2 = 6'2 + c^ - 2bd. (1)
But - d = c cos A.
Therefore a^ = b^ + c'^ - 2bc cos A, (2)
Likewise we learn from geometry that the square of any side oppo-
site an obtuse angle of an oblique triangle is equal to the sum of the
squares of the other two sides increased by twice the product of one of
21
322 ELEMENTARY MATHEMATICAL ANALYSIS (§172
those sides by the projection of the other on it. Thus, in Fig. 131 (2) ,
a2 = 62 + c2 + 2bd (3) '
But d = c cos DAB = c cos (180 — A) = — c cos A.
Therefore (3) becomes
a2 = b'' + c^ - 2bc cos A. (4)
This is the same as (2), so that the trigonometric form of the geo-
metrical theorem is the same whether the side first named is oppo-
site an acute or opposite an obtuse angle.
In the same way we may show that, in any^ triangle
b2= c2-|-a2 -2cacosB, (5)
c2 = a^-l-b^ - 2ab cosC. (6)
Independently of the theorem from plane geometry, we note
from Fig. 131 (1)
a^ = (b - dy -f- p2 = (6 _ dy + c'' - d^
= 62 + (.2 _ -2,bd
= fe'' -h c^ - 26c cos A.
From 131 (2)
o" = (6 -I- dy + p2 = (6 -h dy -I- c2 - d^
= 6=' -I- c2 -I- 2bd
= 62 + 0^-1- 26c cos DAB
= 62-)-c' - 26c cos A,
since DAS = 180° - A and cos (180° - A) = - cos A.
Second Phoop: Since any side of an oblique triangle is
the sum of the projections of the other two sides upon it, the
angles of projection being the angles of the triangle, we have
a = b cos C -|- c cos B,
b = c cos A -f- a cos C, (7)
c = a cos B -h b cos A.
Multiply the first of these equations by a, the second by 6,
the third by c, and subtract the second and third from the first.
The result is
a^ — b^ — c^ — ah cos C -\- ca cos B
— 6c cos A — ab cos C
— ca cos B — be cos A
= — 26c cos A,
or a^ = 6^ -h c'' — 26c cos A.
§173]
TRIGONOMETRIC EQUATIONS
323
173. Law of Tangents. An important relation results if we
take formula (5) §171 by composition and division. First
write the law of sines in the form
sin A
sin jB'
(1)
Then, by composition and division, the sum of the first anteced-
ent and consequent is to their difference as the sum of the second
antecedent and consequent is to their difference; that is
a + h _ sin A + sin B, ,„,
a — & sin A — sin B
Expressing the sums and difference on the right-hand side of (2) as
products by means of the formulas (5) and (6) of §167, we obtain
a + b ^ 2 sin i(,A + B) cos i(A - £)',
a-h 2 cos K^ + B) sin |(A - B)
or simplifying and replacing the
ratio of sine to cosine by the tan-
gent, we obtain
(3)
a-l-b ^ tan \{k + B).
a - b tan J(A - B)
In like manner it follows that
b 4- c tan |(B 4- C).
b - c tan KB - C)
0 4- a _ tan_|(C_+A)
0 — a tan KC — A)
(4)
Fig. 132. — Geometrical
(R\ derivation of law of tan-
gents.
Expressed in words: In any triangle, the sum of two sides is to
their difference, as the tangent of half the sum of the angles opposite
is to the tangent of half of -their difference.
Geometrical Proof: From any vertex of the triangle as center,
say C, draw a circle of radius equal to the shortest of the two sides of
the triangle meeting at C, as in Fig. 132. Let the circle meet the side
a&t R and the same side produced at E. Draw AE and AR. Call
the angles at A, a, and /3, as shown. Then BE = a + 6 and
BR = a - b. Also
a + P = A,
324 ELEMENTARY MATHEMATICAL ANALYSIS [§174
and /. CRA = 0 + B (the external angle of a triangle RAB is equal to
the sum of the two opposite interior angles), or
a - 0 = B.
Therefore
a = lU + B),
P = i{A- B).
Draw RS \\ to EA. ZEAR = ZARS = 90°.
By similar triangles
BE/BR = AE/'SR
^AE . 8R
AR ■ AR
But BE = a + b and BR = a - b, while
AE . ,SR.„
-r-p- = tan a and -j-k = tan p.
AR AR
Therefore a+6 ^ tan KA + -B).
inerelore ^ _ ^ ^^^ ^^^ _ ^^
174. The following special formulas are readily deduced from the
sine formulas and are sometimes useful as check formulas in computa-
tion. They are closely related to the law of tangents. From the
proportion '
a:b:c = sin A: sin B:sin C
by composition ,
c _ sin C
0+6 sin A + sin B
Now by §165 (1) and §167 (5) this may be written
c 2 sin jC cos jC
0 + 6 ~ 2 sin UA + B) cos i(A - B)'
Since C = 180° - (A + B), therefore
C/2 = 90° - |(A + B), and cos C/2 = sin UA + B).
c sin iC cos |(A + B) ,,,
a + b cos i(A — B) cos i(A — B;
In like manner it can be proved that
c _ sin i(A + B) ,2)
a - b sin |(A - B)
Both (1) and (2) can be readily deduced geometrically from Fig. 132.
176. The s-formulas. The cosine formula
y
a2 = 62 -}- c^ - 26c cos A
§175] TRIGONOMETRIC EQUATIONS 325
can be written in the forms
a^ = (b + cy - 26c(l + cos A), (1)
a2 = (6 - c)2 + 26c(l - cos A), (2)
by adding (+26c) and (—26c) to the right-hand member in each
case. But now we know from §166, (1) and (2), that
1 + C0S A = 2 cos" (A /2),
1 - cos JL = 2 sm" (A/2). ,
Therefore (1) and (2) above become
o= = (6 + c)2 - 46c cos" (A/2), (3)
' a" = (6 - cy + 46c sin" (A/2). (4)
Writing these in the form i\i
ibc sin" (A/2) = o"]- (6 - c^, (5)
46c cos" (A/2) = (6 + c)" - a", (6)
and dividing the members of (5) by the members of (6), we
obtain
tan" (A/2) =1^1^. (7)
Factoring the numerator and denominator we obtain
tan" (A/2) -fi + ^Tu^T^ + l- (8)
'(6 + c + o) (6 + c — a)
Let the perimeter of the triangle be represented by 2s, that is,
let
a + 6 + c = 2s.
Hence, subtracting 2c, 26, and 2a in turn,
a + 6 — c = 2s — 2c (subtracting 2c),
a — 6-|-c = 2s — 26 (subtracting 26),
6 + c — a = 2s — 2a (subtracting 2a).
Therefore equation (8) becomes
tan" (A/2) = (^ -/>)i^ ' <=) . (g)
s(s — a)
Let
(s - a) (s - 6) (s - c) /s = r\ (10)
326 ELEMENTARY MATHEMATICAL ANALYSIS [§175
Then
or '
Likewise
tan^ (A/2) = r-V(s - aY,
tan (A/2) = r/(s — a).
tan (B/2) = r/(s - b),
tan (C/2) = r/(s - c),
(11)
(12)
(13)
Fig. 133. — Geometrical derivation of the s-formulas.
Geometrically: These formulas may be found by means of the
diagram Fig. 133. Let the circle 0 be inscribed in the triangle ABC;
its center is located at the intersection of the bisectors of the internal
angles of the triangle. Let its radius be r. ATi = ATt, BTt = BTz,
CTi = CTi, and since 2s = a + 6 + c, it follows that one way of
writing the value of s is
s = BTi + TiC + ATi.
§176] TRIGONOMETRIC EQUATIONS 327
Therefore
ATi = s -a.'
Hence it follows that
tan (A/2) = r/(s - o). (14)
Since this result is the same as (11) above, it proves that the r of
equation (10) is the radius of the inscribed circle, and therefore proves
that the radius of the inscribed circle may be expressed by the formula
Us - a)is -6)(s -c)
a fact that is usually proved in text books on plane geometry.
176.* Miscellaneous Formulas for Oblique Triangles. The fol-
lowing formulas are given without proof. They are occasionally
useful for reference, although no use will be made of them in
this book. The following notation is used: The three sides of
the oblique triangle are named a, b, c, and the angles opposite
these A, B, C, respectively. The semi-perimeter of the triangle
is s, OT 2s = a + b + c. The radius of the circumscribed circle
is B, that of the inscribed circle is r, and the radii of the escribed
circles are Ta, n, r^ tangent, respectively, to the sides a, b, c
of the given triangle. K stands for the area of the triangle.
s = 4i? cos iA cos J-B cos §(7 (1)
s — c — 4Rsin iA sin ^B cos iC (2)
and analogs for s — a and s — b.
r = iR sin JA sin iB sin iC (3)
Tc = 4jB cos iA cos iB sin iC (4)
and analogs for Ta and rt.
Ta = s tan iA,n = s tan iB, r^ = s tan JC (5)
2K = ab sinC = be sin A = ca sin B (6)
K = 2R'' sin A sin S sin C = |^ (7)
K = Vsis -a) is- b) (s - c) (8)
K = rs = ra(s — a) = n(s — 6) = r^is — c) (9)
Z2 = rr^nr, (10)
K^ = {s - a) tan iA = {s - b) tan iB = {s - c) tan |C (11)
328 ELEMENTARY MATHEMATICAL ANALYSIS [§177
C. NUMERICAL SOLUTION OF OBLIQUE TRIANGLES
177. An oblique triangle possesses six elements; namely, the
three sides and the three angles. If any three of these six
magnitudes be given (except the three angles), the triangle is
determinate, or may be constructed by the methods explained
in plane geometry; it will also be found that if any three of these
six magnitudes be given, the other three may be computed by the
formulas of trigonometry, provided, that the given parts include
at least one side.
It is convenient to divide the solution of triangles into four
cases, as follows :
I. Given two angles and one side.
II. Given two sides and an angle opposite one of them.
III. Given two sides and the included angle.
IV. Given the three sides.
The solution of these cases with appropriate checks will now
be given. The best arrangement of the work of computation
usually consists in writing the data and computed results in the
left margin of a sheet of ruled letter paper (SJ inches X 11 inches)
and placing the computation in the body of the sheet. Every
entry should be carefully labeled and computed results should be
enclosed in square brackets. AU work should be done on ruled
paper and invariably in ink. Special calculation sheets (forms
M2 and M7) have be'en prepared for the use of students. Neat-
ness and systematic arrangement of the work and proper checking
are more important thanr rapidity of calculation.
178. Computer's Rules. The following computer's rules are
useful to remember in logarithmic work :
Last Digit Even: When it becomes necessary to discard a
5 that terminates any decimal, increase by unity the last digit
retained if it be an odd digit, but leave it unchanged if it be an
even digit; that is, keep the last digit retained even. Thus log tt
= 0.4971; hence write | log x = 0.2486. Also log sin 18° 5'
= 9.4900 + (correction) 19.5 = 9.4920.
Of course if the discarded figure is greater than 5, the last
digit retained is increased by 1, whUe if the discarded figure is
less than 5, the last digit retained is unchanged.
8179]
TRIGONOMETRIC EQUATIONS
329
Functions or Angles in Second Quadbant: In finding
from the table any function of an angle greater than 100° (but
< 180°) replace the first two digits of the number of degrees in the
angle by their sum and take the co-function of the result. The
method is valid because it is equivalent to the subtraction
of 90° from the angle. By §163 this always gives the cor-
rect numerical value of the function. The algebraic sign should
be taken into account separately. Thus, sin 157° 32' 7" =
cos 67° 32' 7". In case of an angle between 90° and 100°,
ignore the first digit and proceed in the same way. Thus,
tan 97° 57' 42" = - cot 7° 57' 42"
179. Case I. Given two angles and one side, as A, B, and c.
1. To find €, use the relation A+B + C = 180°.
2. To find a and 6, use the law of sines, §171.
3. To check results, apply the check formula (1) or (2) §174.
Illtjstkation: In an oblique triangle, let c = 1492, A = 49° 52',
B = 27° 15'. It is required to compute C, a, b.
The following form of work is self explanatory. It should be noted
that the process of work and the meaning of each number entering the
calculation is properly indicated or labeled in the work.
Numerical Work
Given
To find o, b, and C.
c = 1492
Formulas
A = 49° 52'
C = 180 - U + B) =
= (102° 53')
B = 27° 15'
c sin A
Work.
" ~ sin C
, _ c sin B
sin C
log sin A =
log c =
9.8834 -
3.1738
■ 10
log sin B =
9.6608 -
- 10
^
log sin C =
9.9889 -
- 10
log o
3.0683
log 6
2.8457
a =
[1170.]
6
[701.]
Check.
Check Formula
c -b =
: 791
a sin i(C + B)
C + B =
: 130° 8'
c
- h sin i(C - B)
C - B ^
: 75° 38'
330 ELEMENTARY MATHEMATICAL ANALYSIS [§I80
i(C + B) = 65° 4'
UC - B) = 37° 49'
log a = 3.0683
log c - 6 = 2.8982
log r^b = 01701
log sin i {C + B) = 9.9575 - 10
log siD i (e - B) = 9.7875 - 10
Examples
Find the remaining parts, given :
1. A = 47° 20', B = 32° 10',
2. B = 37° 38', C = 77° 23',
3. B = 25° 2', C = 105° 17',
4. C = 19° 35', A = 79° 47',
Check
a = 739.
6 = 1224.
6 = 0.3272.
c = 56.47.
180. Case n. Given two sides and an angle opposite one of
hem, as a, b, and A .
Fig. 134. — Case II of triangles, for one, two, and impossible
solutions.
1. To find B, use the law of sines, §171.
2. To find C, use the equation A +B + C = 180°.
3. To find c use the law of sines.
4. To check, apply the check formula (I) or (2), §174.
When an angle as B, above, is determined from its sine, it admits
of two values, which are supplementary to each other. There
may be, therefore, two solutions to a triangle in Case II. The
solutions are illustrated in Fig. 134.
§180]
TRIGONOMETRIC EQUATIONS
331
• In case one of the two values of B when added to the given
angle A gives a sum greater than two right angles, this value
of B must be discarded, and but one solution exists. If a be
less than the perpendicular distance from C to c, no solution
is possible.
Illustration: Solve the triangle if a = 345, 6 = 534, and
A = 25° 25'.
The solution is readily understood from the following work.
Numerical Work
Given
a = 345
6 = 534
To find c, B, and C.
Formulas
. „ b sin A
A = 25° 25'
a
Work.
C = 180 - (A + B)
log 6 =2.7275
a Bin C
log sin jl = 9.6326 - 10
sin A
log a =2.5378
logsmB = 9.8223 - 10
B = [41°37'.l]
or
[138° 22'. 9]
A+B =67°2'.l
163° 47'. 9
C = [112° 57'. 9]
or
[16°12'.l]
logo =2.5378
2.5378
log sin C = 9.9641 - 10
9.4456 - 10
log sin A = 9.6326 - 10
9.6326 - 10
log c = 2.8693
2.3508
c = [740.1]
[224.3]
Check
6 sin i(C + A) ^
a _ sin K-B + C)
c — a sm t
c — a
C + A
C — A
W + A)
KC - A)
log 6
log (c - o)
logQ
log sin KC +
log sin KC —
logO
iC -A)
= 395.1
= 138° 22'.
= 87° 32'.
= 69° 11'
= 43° 46'
= 2.7275
= 2.5967
= 0.1308
= 9.9707 -
A) = 9.8400 -
= 0.1307
A)
10
10
b — c sin |(B
b -c
B + C
B -C
UB + C)
i(B - C)
log o
log (6 - c)
logQ'
Iogsini(B + C)
logsini(fi-C)
logQ'
= 309.7
= 154° 35'
= 122° 10'. 8
= 77° 17'. 5
= 61° 5'. 4
= 2.5378
= 2.4910
= 0.0468
= 9.9892 -
= 9.9422 -
= 0.0470
10
10
332 ELEMENTARY MATHEMATICAL ANALYSIS [§181
Examples
Compute the unknown parts in each of the following triangles :
, 1. a = 0.8, b = 0.6, B = 40° 15'.
2. o = 8.81, 6 = 11.87, A = 19° 9'.
3. 6 = 81.05, c = 98.75, C = 99° 19'.
4. c = 50.37, a. = 58.11, C = 78° 13'.
6. a = 1213, 6 = 1156, B = 94° 15'.
181. Case III. Given two sides and the included angle, as a,b,C.
1. To find A +B,useA +B = 180° - C.
2. To find A and B, compute (A — B)/2 by the law of tangents,
§173, equation (4), then A = (A + B)/2 + (A - B)/2 and
B = (A+ B)/2 - {A - B)/2.
3. To find c, use law of sines, §171.
4. To check, use law of sines.
Illustration: Given a = 1033, 6 = 635, C = 38° 36'.
Numerical Work
Given
To find c. A, and B.
a =
1033 Formulas
b =
635 A+B = 180 -C = 141° 24'
C =
3*° 3^' tan UA-B) = ^-=-^ tanJCA + B)
o'sin C
tsin A
Work
a -b ' = 398
a + b = 1668
\{A+B) = 70° 42'
log (a - 6) = 2.5999
logtanKA+'B) =0.4557
- log(o + 6) =3.2222
log tan 4U -B) = 9.8334 - 10
1{A -B) = 34° 16.3'
A =[104° 58.3']
B = [36° 25.7']
logo =3.0141
log sin C = 9.7951 - 10
log sin A = 9.9850 - 10
logo =2.8242
c = [667.1]
§182]
TRIGONOMETRIC EQUATIONS
333
Check
b sin C
sin B
log 6 =2.8028
log sin C = 9.7951 - 10
log sin B = 9.7737 - 10
logc =2.8242
c = [667.1]
Examples
Compute the unknown parts in each of the following triangles :
1. a =78.9, 6=68.7, C = 78° 10'.
2. c = 70.16, a = 39.14, B = 16° 16'.
3. 6 = 1781, c = 982.7, A = 123° 16'.
182. Case IV. Given the three sides.
1. To find the angles, use the s-formulas, §175, (11),
and (13).
2. To check, use A + B + C = 180°.
Illustration: Given a = 455, 6 = 566, c = 677, find A, B and C.
Numerical Work
(12)
Given
"Work.
a =
455
6 =
566
c =
677
2s =
1698
s =
849
s
— a =
394
s
-6 =
283
s
— c =
172
To find A, B and C.
Formulas
T
tan iA = '
tan hB =
tan JC =
s - b
r
where r = ^'(^ - aKs - bHs - e).
2si = 1698
log (s-a) = 2.5955
log (s-b) = 2.4518
log (s - c) = 2.2355
logs =2.9289
logr" =4.3539
logr =2.1770
log tan JA = 9.5815 - 10
log tan IB = 9.7252 - 10
log tan iC = 9.9415 - 10
• Adding th« four numbera above this line cheoks the subtractions (> — a),
{» -h), etc.
334 ELEMENTARY MATHEMATICAL ANALYSIS (§182
iA = 20° 53'
JJS = 27° 58'
JC = 41° 9'
A = [41° 46']
B = [55° 56']
C = [82° 18']
Check.
A+B+C = 180°
Exercises
Find the values of tlJe angles in each of the following triangles :
1. a = 173, 6 = 98.6, c = 230.
2. a = 8.067, 6 = 1.765, c = 6.490.
3. a = 1911, 6 = 1776, c = 1492.
Miscellaneous Problems
The instructor will select only a limited number of the following
problems for actual computation by the student. The student
should be required, however, to outline in writing the solution of a
number of problems which he is not required actually to compute, and,
when practicable, to block out a suitable check for each one of them.
1. From one corner P of a triangular field PQB the side PQ bears
N. 10° E. 100 rods. QR bears N. 63° E. and PR bears N. 38° 10' E.
Find the perimeter and area of the field.
2. The town B lies 15 miles east of A, C lies 10 miles south of A.
X lies on the Hne BC, and the bearmg of AX is S. 46° 20' E. Find
the distances from X to the other three towns.
3. To find the length of a lake (Fig. 135), the angle C = 48° 10', the
side a = 4382 feet, and the angle B = 62° 20' were measured. Find
the length of the lake c, and check.
4. To continue a line past an obstacle L, Pig. 136, the line BC and
the angles marked at B and C were measured and found to be 1842
feet, 28° 15', and 67° 24', respectively. Find the distance CD, and
the angle at D necessary to continue the line AB; also compute the
distance BD.
5. Find the longer diagonal of a parallelogram, two sides being
69.1 and 97.4 and the acute angle being 29° 34'.
What is the magnitude of the single force equivalent to two forces
of 69.1 and 97.4 dynes respectively, making an angle of 29° 34' with
each other?
6. A force of 75.2 dynes acts at an angle of 35° with a force F.
Their resultant is 125 dynes. What is the magnitude of Fl
§182]
TRIGONOMETRIC " EQUATIONS
335
7. The equation of a circle is p = 10 cos 6. The points A and
B on this circle have vectorial angles 31° and 54° respectively. Find
the distance AB, (1) along the chord; (2) along the arc of the circle.
8. Knd the lengths of the sides of the triangle enclosed by the
straight lines :
e = 26°
115°; p cos (9 - 45°) = 50.
Fig. 135. — Diagram for
Problem 3.
Fig. 136. — Diagram for Problem 4.
9. A gravel heap has a rectangular base 100 feet long and 30 feet
wide. The sides have a slope of 2 in 5. Find the number of cubic
yards of gravel in the heap.
10. A point B is invisible and inaccessible from A and it is necessary
to find its distance from A. To do this a straight line is run from A
to P and continued to Q such that B is visible from P and Q. The
following measurements are then taken: AP, = 2367 feet; PQ = 2159
feet; APB = 142° 37'.3; AQB = 76° 13'.8. Find AB.
11. To determine the height of a mountain the angle of elevation
of the top was taken at two stations on a level road and in a direct
line with it, the one 5280 yards nearer the mountain than the other.
The angles of elevation were found to be 2° 45' at the further station
and 3° 20' at the nearer station. Find the horizontal distance of the
mountain top from the nearer station and the height of the mountain
above it. Use S and T functions.
12. Explain how to find the distance between two mountain peaks
Ml and Af 2, (1) when A and B at which measurements are taken are in
the same vertical plane with Mi and M^; (2) when neither A nor B
is in the same vertical plane with Mi and M2.
13. The sides of a triangular field are 534 yards, 679 yards, and 474
yards. The first bears north, and following the sides in the order here
given the field is always to the left. Find the bearing of the other
two sides 'and the area.
336 ELEMENTARY MATHEMATICAL ANALYSIS [§182
14. From a triangular field whose sides are 124 rods, 96 rods, and
104 rods a strip containing 10 acres is sold. The strip is of uniform
width, having as one of its parallel sides the longest side of the field.
Knd the width of the strip.
16. Three circles are externally mutually tangent. Their radii are
5, 6, and 7 feet. Find the area and perimeter of the three-cornered
area enclosed by the circles and the
length of a wire that will enclose the
group of three circles when stretched
about them.
16. To find the distance between two
inaccessible objects C and D, Kg. 137,
two points A and B are selected from
which both objects are visible. The dis-
137_ Diagram for tS'^ce AB is found to be 7572 feet.
Problem 16. The following angles were then taken:
ABD = 122° 37' BAC = 80° 20'
ABC = 70° 12' BAD = 27° 13'
Knd the distance DC and check.
17. A circle of radius o has its center at the point (pi, 9i). Knd its
equation in polar coordinates. (Use law of cosines.)
18. A surveyor desired the distance of an inaccessible object 0
from A and B, but had no instruments to measure angles. He
measured AA' in the Une AO, BB' in the line BO; also AB, BA', and
AB'. How did he find OA and OB?
19. From a point A a distant object C bears N. 32° 16' W. and
from B the same object bears N. 50° W. AB bears N. 10° 39' W.
The distance AB is 1000 yards. Knd the distance AC.
20. The angle of elevation of a mountain peak is observed to be
19° 30'. The angle of depres.sion of its image reflected in a lake 1250
feet below the observer is found to be 34° 5'. Find the height of the
mountain above the observer and the horizontal distance to it. (See
Fig. 138.)
21. One side of a mountain is a smooth eastern slope inclined at an
angle of 26° 10' to the horizontal. At a station A a vertical shaft is
sunk to a depth of 300 feet. From the foot of the shaft two horizontal
tunnels are dug, one bearing N. 22° 30' E. and the other S. 65° E.
These tunnels emerge at B and at C respectively. Find the lengths
of the tunnels and the lengths of the sides of the triangle ABC.
22. A rectangular field ABCD has side AB = 40 rods; AD = 80
rods. Locate a point P in the diagonal AC so that the perimeter of
the triangle APB will be 160 rods. {Hint: Express perimeter as a
function of angle at P.)
§182]
TRIGONOMETRIC EQUATIONS
337
8. Find the area enclosed by the lines y = k' y = \/3 x, and the
Fig. 138. — Diagram for Problem 20.
circle x' — lOs + ^^^ = 0. (Hint: Change to polar coordinates.)
24. The displacement of a particle from a fixed point is given by
d = 2.5 cos t + 2.5 sin t.
What values of t give maximum and minimum displacements; what
is the maximum displacement?
25. A quarter section of land is enclosed by a fence. A farmer
wishes to make use of this fence and 60 rods of additional fencing in
making a triangular field in one comer of the original tract. Find
the field of greatest possible area. Show that it is also the field of
maximum perimeter, under the conditions given.
26. A force Fi = 100 dynes makes an angle of 6° with the horizontal,
and a second force Fi = 50 dynes makes an angle of 90° with Fi.
Determine B so that (1) the sum of the horizontal components of
Fi and Ft shall be a maximum; (2) so that the sum of the vertical com-
ponents shall be zero.
27. Find the area of the largest triangular field that can be enclosed
by 200 rods of fence, if one side is 70 rods in length.
22
338 ELEMENTARY MATHEMATICAL ANALYSIS [§182
28. Change the equation of the curve xy = I to polar coordinates,
rotate through — 45°, and change back to rectangular coordinates.
29. A particle moves along a straight line so that the distance
varies directly as (sin t + cos t). When t = 7r/4, the distance is 10.
Find the equation of motion.
30. From the top of a lighthouse 60 feet.high the angle of depression
of a ship at anchor was observed to be 4° 52'; from the bottom bf the
lighthouse the angle was 4° 2'. Required the horizontal distance from
the lighthouse to the ship and the height of the base of the lighthouse
above the sea.
31. The Une AB runs north and south. The line AC makes an
angle of 52° 8'. 6 with AB. Locate the Une BC perpendicular to AB
so that the area ABC shall be 1 acre.
32. University Hall casts a shadow 324 feet long on the hillside
on which it stands. The slope of the hillside is 15 feet in 100 feet,
and the elevation of the sun is 23° 27' Find the height of the building.
33. To determine the distance of a fort A from a place B, a line BC
and the angles ABC and BCA were measured and found to be 3225.5
yards, 60° 34', and 56° 10' respectively. Find the distance AB.
34. A balloon is directly over a straight level road, and between two
points on the road from which it is observed. The points are 15,847
feet apart, and the angles of elevation are 49° 12' and 53° 29'. Find
the height.
35. Two trees are on opposite sides of a pond. Denoting the trees
by A and B, we measure AC = 297.6 feet, BC = 864.4 feet, and the
angle ACB = 87° 43'. Find AB.
36. Two mountains are 9 and 13 miles respectively from a town,
and they include at the town an angle of 71° 36'. Find the distance
between the mountains.
37. The sides of a triangular field are, in clockwise order, 534 feet,
679 feet, and 474 feet; the first bears north; find the bearings of the
other sides and the area.
38. Under what visual angle is an object 7 feet long seen when the
eye is 15 feet from one end and 18 feet from the other?
39. The shadow of a cloud at noon is cast on a spot 1600 feet west
of an observer, and the cloud bears S., 76° W., elevation 23°. Find
its height.
CHAPTER XI
SIMPLE HARMONIC MOTION AND WAVES
183. Simple Harmonic Motion. In Fig. 139, x = 0T> =
a cos DOM, where a is the radius of the circle. If now the point
M is thought of as moving with constant or uniform speed on the
circle, starting at A, or (which amounts to the same thing) if
the radius OM is thought of as moving with constant angular
velocity, say k radians per second, starting from OA, then angle
DOM = kt and the position of the point D at time t is given by
X = a cos kt, (1)
where t is the time in seconds required for OM to move from posi-
tion OA to position OM.
Let us study the motion of the point D as M moves on the
circle with constant speed.
D starts at A and moves to
the left with increasing speed
until it arrives at ■ 0, where
its speed begins to decrease,
decreasing to 0 at A'. Then
the point moves to the right
with increasing speed until it
again passes through 0, after
which its speed diminishes,
becoming 0 when it arrives
at A . Then the whole motion
is repeated. A body whose
position on a straight line is
given at any instant by an equation of the form (1), that is one
which moves as the point D does, is said to describe simple harmonic
motion. On account of the frequency with which this term will
occur, we shall abbreviate it by the symbols S.H.M. Examples
.339
Fig. 139.
340 ELEMENTARY MATHEMATICAL ANALYSIS [§183
of bodies that move approximately in this way are: The bob of a
pendulum, a point in the prong of a vibrating tuning fork, a
point in a vibrating violin string, the particles of air during the
passage of a sound wave. The motion is oscillatory in character
and repeats itself in definite intervals of time.
The length of this interval can be easily found by considering
the motion of the point M on the circle. The point D starting
from any given position will return to this position moving in the
same direction after an interval of time which is the time required
for M to describe the circle, i.e., after 2ir/k seconds, the time in
which the radius OM describes the angle 2ir radians at the rate of
k radians per second. This time within which a body executing
S.H.M. performs a complete oscillation is called the period of the
S.H.M. It is denoted by T. Thus
T = ^. (2)
This expression can be obtained directly from the equation x =
a cos kt by means of the fact that the cosine is a periodic function
of period 2x. The period T is the amount by which t must be
increased in order to increase the angle kt by the amount 27r.
If t be increased by the amount 2ir/k, then kt is increased by
2x, because
fc(t+^) =A;i + 2ir.
The number of complete periods per second is
^ = T = .V (3)
N is called the frequency of the S.H.M.
'if instead of counting time from the instant at which the
auxiliary point M passed through A, we count it from the instant
it passed through E, then ZEOM = kt, and it is clear that
ZAOM = (kt — e) if e stands for the constant angle EOA.
Then (1) becomes
X = a cos (Jet — «). (4)
The number a is called the amplitude, e is called the epoch angle,
§184] SIMPLE HARMONIC MOTION AND WAVES 341
and (Jet — e) is called the phase angle of the S.H.M. represented
by (4).
In like manner the point D2, the projection of the point M upon
the vertical diameter of the circle in Fig. 139, describes S.H.M.
Its equation is
2/ = a sin (kt — e), (5)
where time t is measured from the instant M passes through E.
184. Mechanical Generation of S.H.M. Fig. 140 illustrates
a way in which S.H.M. may be described by mechanical means.
rp
^
-
B
—3" \r\\
^
^.
<
^4^^ ho.
C K
1
^
\
2?__"5 =__
> s'
—L
7L S
'^ : - -
- ,7 _ _ _ _ s_ .
'-'-1
s,
7 ■S
s :, ,
^ €. S .,
_ Si^: cj-
7 _S ^1
\
ii
_^
c 1 S '
' ^
~-~.
■s ?
v^
"5 7'
\? rs M
"--III II llml II 1 WyW II
H
■^
|LJ_LiJ_LJ_LJ_ri I ^ 1
Pig.
140. — Mechanical generation of simple harmonic motion,
and of a simple progressive wave.
Let the uniformly rotating wheel OAB be provided with a pin
M attached to its circumference and free to move in the slot
of the cross-head as shown, the arm "of the cross-head being re-
stricted to vertical motion by suitable guides G-G\- Then, as the
wheel rotates, any point P of the arm of the cross-head describes
simple harmonic motion in a vertical direction. The amplitude
of the S.H.M. is the radius of the circle, or OB; its period is the
time required for one complete revolution of the wheel.
342 ELEMENTARY MATHEMATICAL ANALYSIS [§18S
Exercises
1. Find the periods of the following S.H.M. :
{a) y = 3 sin 2t. (e) y = a sin (10< — 7r/3).
(6) 2/ = 10 sin (1/2) « (/) t/ = o sin (2«/3 - 27r/5).
(c) y = 7 cos 4<. ig) y = a sin (6< + c).
{d) y = a sin 27r<.
2. Give the amplitudes and epoch angles in each of the instances
given in exercise 1.
3. The bob of a second's pendulum swings a maximimi of 4 cm.
each side of its lowest position. Considering the motion as rectilinear
S.H.M., write the equation of motion.'
^ The term period is used differently in the case of a pendulum than in the case
of S.H.M. The time of a swing is the period of a pendulum; the time of a awino-
swang is the period of a S.H.M.
Write the equation of motion of a pendulum of the same length
which was released from the end of its swing 1/2 second after the first
pendulum was similarly released.
4. A particle moves in a straight Une in such a way that its dis-
placement from a fixed point of the line is given by d = 2 cos* t. Show
that the particle moves in S.H.M., and find the amplitude and period
of the motion.
6. A particle moves in a vertical circle of radius 2 units with angular
velocity of 20 radians per second. Counting time from the instant
the particle was at its lowest position, write the equation of motion
of its projection (1) upon the vertical diameter; (2) upon the horizon-
tal diameter; (3) upon the diameter bisecting the angle between the
horizontal and vertical.
186. S.H.M. Record on Smoked Glass. If P, Fig. 140, be a
tracing point attached to the vertical arm of the cross-head and
capable of describing a curve on a piece of smoked glass, HK, which
is moved to the right at constant speed, then when P describes
S.H.M. in the vertical line OP, the curve NiCTNJ' traced on the
plate HK is a sinusoid. For, if iVj be taken as origin, and if for
convenience positive abscissas be measured to the left, the coordi-
nates of P are
X = Vt,
and y = a sin {kt — e)
§186] SIMPLE HARMONIC MOTION AND WAVES 343
where V is the linear velocity of the plate. Eliminating t
between these two equations,
y = asin yy - ej (1)
the equation of a sinusoid.
If the plate HK moves with the same velocity as the point M,
we have
V = ha
and equation (1) becomes
- = sm -, (2)
a a
the equation of an undistorted sinusoid.'
186.* Composition of Two S.H.M.'s at Right Angles.
It is obvious that
X = a cos ht
represents a S.H.M. one quarter of a period in advance of a;' = a sin kt,
since sin I fci + ^1 = cos kt. A pair of S.H.M.'s possessing this
property are said to be in quadrature. (4) and (5), §183, may be said
to be in quadrature.
We have shown that if a point M, moving uniformly on a circle, be
projected upon both the X- and 7-axes, two S.H.M.'s result. The
phase angles of these two motions differ from each other by 7r/2.
The converse of this fact, namely that uniform motion in a circle may
be the resultant of two S.H.M.'s in quadrature, is easily proved, for the
two equations of S.H.M.
X = a cos kt
y = a sin kt
are obviously the parametric equations of a circle. Hence the theorem :
Uniform motion in a circle may he regarded as the residtant of two
S.H.M.'s of equal amplitudes and equal periods and differing by 7r/2 in
phase angle.
This important truth is illustrated by Fig. 141. Let the X- and
1 The student should note that ^ = sin - is of exactly the same shape as y = sm x,
for multiplying both ordinates and abscissas of any curve by a is merely constructing
the curve to a different scale. However, ^ = sin o is a distorted sinusoid, for the
ordinates of y = sin x are multiplied by 3 while the abscissas are multiplied only
by 2.
344 ELEMENTARY MATHEMATICAL ANALYSIS [§187
y-axes be divided proportionally to the trigonometric sine, as in Fig.
59. Through the points of division of the two axes draw lines per-
pendicular to the axes, thus dividing the plane into a large number of
small rectangles. Starting at the end of one of the axes, and sketch-
ing the diagonals of successive cornering rectangles, the circle ABA'B'
is drawn.
If the same construction be carried out for the case in which the Y-
axis is divided proportionally to 6 sin kt and in which the X-axis is
divided proportionally to osin
kt, the ellipse AiBiA'iB'i re-
sults. These facts are merely
a repetition of the statements
made in §84.
187. Waves.— Let Fig. 142
represent a section obtained
by passing at any instant a
vertical plane perpendicular
to the crests of a series of
small waves on the surface
of a body of water. The
wavy line represents the ap-
pearance of the surface at
any instant. It is a fact
that its equation is, in the
case of small waves or ripples,
■
1 ,
B
--
-7-
—
' —
—
—
^s-::
7
^ S
/
S
/
^
,
0
\
\
1
\
\
7
A
7 _
.__
—
—
—
—
^2-:
■p
ff\r
—
—
—
^
V--S-
— F
—
—
—
—
g — 5,
n'
1
Ai --
[
1
,
■-:
. = -^
1
1
i--^-^-
Ax
2/ = a sin he.
(1)
Fig. 142 represents the seo-
■^1 tion of the surface at any
Fig. 141. — The circle and the instant, say t = 0. Now
eUipse considered as generated by „„„i „,„„„„ „„„„ c a
two S.H.M.'s in quadrature. ^'^.^f ^*^«3 °iove forward
with a constant velocity,
which we shall call Y. The wavy form is sinusoidal in section
but of course it is not fixed, but keeps moving ahead. Hence
the moving sinusoid of Fig. 140 may be looked upon as a repre-
sentation of this kind of phenomena.
The curve described on the moving plate UK of Fig.- 140, if
referred to coordinate axes moving with the plate, is the sinusoid, or
§187] SIMPLE HARMONIC MOTION AND WAVES 346
sine curve, whose equation is (1) above. If, however, we consider
this curve as referred to the fixed origin Oi, then the moving
sinusoid thus conceived is called a simple progressive sinusoidal
wave or merely a wave. Under the conditions represented in
Fig. 140, it is a wave progressing to the right with the uniform
speed of the plate HK. At any single instant, the equation of
the curve is
y = a sin h{x - OiN), (2)
where OiN is the distance that the node N has been translated
to the right of the origin Oi. If V be the uniform velocity of
translation of HK, then,
OiN = Vt (3)1
Fig. 142.
and the equation of the wave is
y = asiah{x — Vt),
or y = a sin Qix — hVt),
or y = a sin (hy — kt),
if k be put for hV, so that
V = -
(4)
(5)
Because of the presence of the variable t, (4) is not the equation
of a fixed sinusoid, but of a moving sinusoid or wave.
Applying the same terms used fbr S.H.M., the expression
{hx — ht) is the phase angle, the expression (+ kt) is the epoch
angle and a is the amplitude of the wave. See Fig. 143a and c.
The expression Qix — kt) is a linear function of the variables
1 In what follows, t is not the time elapsed since itf , Fig. 140, was at A, as used in
S183| but is the elapsed time since N was at 0i, These values of t differ by the time
of half a revolution or by ir/k.
346 ELEMENTARY MATHEMATICAL ANALYSIS [§187
X and t. The sine or cosine of this function is called a simple
harmonic fmiction of x and t.
The wave form on the surface of water moves along with fixed
velocity V. The particles of water, however, do not share in this
b
^''S<^^'yr>
<.
X X X/
Fig. 143. — Wave forms, (a) of different amplitude; (5) of different
wave lengths; (c) of different phase or epoch angles.
forward motion. Each particle on the surface moves up and
down in a vertical line as the wave form passes it. In fact we
shaU now see that each particle describes S.H.M. in a vertical
direction.
To examine the motion of a single particle of water, we have
§188] SIMPLE HARMONIC MOTION AND WAVES 347
only to regard x as constant, say x = Xi, in equation (4) above
The displacement of this particle is then given by
y = a sin (hxi — kt)
or y = — a sin (M — hxi).
That is y = a sin {kt — hxi — ir). (6)
This is the equation of a S.H.M. whose period is T = 2T/k. The
epoch angle is hxi + ir. This will be different for different par-
ticles. This means that the phase angles of the S.H.M. of succes-
sive particles differ, but they all oscillate up and down with the
same period 2ir/k.
188. Wave Length. The wave length of a progressive wave is
the distance from crest to crest or from trough to trough. It
is the amount by which x must be increased in the equation of
the wave in order that the angle (hx — kt) may be increased by
2ir. Hence the wave length,
^ = ¥- «
189. Period or Periodic Time. If we fix our attention upon any
particular or constant value of x, and view the progressive wave
as it passes the vertical line through this abscissa, the elapsed
time from the passage of one crest to the next crest is called the
period, or periodic time. It is readily seen to be the increment
in t which changes the angle (hx — kt) by the amount 2ir. Hence
the period
The expression T is called the periodic time, or period, of the
wave. It is the length of time required for the wave to move one
wave length. To contrast wave length and period, think of a per-
son in a boat anchored at a fixed point in a lake. The time that
the person must wait at that fixed point (x constant) for crest to
follow crest is the periodic time. The wave length is the distance
he observes between crests at a given instant of time (t constant) .
The number of periods per unit of time is called the frequency
of the wave. Hence, if N represent the frequency of the wave,
N = |'=|- (2,
348 ELEMENTARY MATHEMATICAL ANALYSIS [§190
190. Velocity or Rate of Propagation. The rate of movement
V of the sinusoid on the plate HK, Fig. 140, is shown by equation
(5), §187, to be k/h units of length per second. This is called the
velocity of the wave or the velocity of propagation. The equa-
tion of the wave may be written
2/ = o sin h{x — Yt).
From equations (1) §188 and (1) §189 we obtain
A;
and since 7 = r, we have
K
k_L
h~ t'
V =^. (1)
This equation is obvious from general considerations, for the
wave moves forward a wave length L in time T, hence the speed
of the wave must be m'
191. L and T Equation of a Wave. If we solve equations (1)
§188 and (1) §189 for h and k respectively, and substitute these
values of h and ifc in the equation
2/ = a sin {hx — kt)
we obtain
■\i-a-
From this form it is seen that the argument of the sine increases
by 2ir when either x increases by an amount L or when t increases
by the amount T. By use of (1), §190, the last equation may
also be written
^- -' - .(2)
a sm2^
y
= asm
L^^-
Vt).
192.
Phase,
Epoch,
Lead.
Consider the two
waves
y
. I27r\
= a sm y- (a; —
Vt)
y
= a sin
2ir .
j-ix-
vt-
E)
a)
(2)
The amplitudes, the wave lengths and the velocities are the
§192] SIMPLE HARMONIC MOTION AND WAVES 349
same in each, but the second wave is in advance of the first by
the amount E (measured in linear units), for the second equation
can be obtained from the first by substituting (x — E) for x, which
translates the curve the amount E in the OX direction. In this
case E is called the lead (or the lag if negative) of the second wave
compared with the first. The lead is a linear magnitude measured
in centimeters, inches, feet, etc.
The terms phase and epoch are sometimes used to designate
the time, or, more accurately, the fractional amount of the period
required to describe the phase angle and epoch angle respectively.
In this use, the phase is the fractional part of the period that has
elapsed since the moving point last passed through the middle point
of its simple harmonic motion in the direction reckoned as positive.
See Fig. 143c.
The tidal wave in mid-ocean, the ripples on a water surface,
the wave sent along a rope that is rapidly shaken by the hand,
are illustrations of progressive waves of the type discuseed above.
Sound waves also belong' to this class if the alternate condensations
and rarefactions of the medium be graphically represented by
ordinates. The ordinary progressive waves observed upon a lake
or the sea are not, however, progressive waves of this type. The
surface of the water in this case is not sinusoidal in form, but
is represented by another class of curves known in mathematics
as trochoids.
Exercises
1. Derive the amplitudOj the wave length, the periodic time, the
velocity of propagation of the following waves :
(a) y = a sin {2x — 3<). , > .„„ . 2w, _.. ..
(b) y =5 sin (0.75a; - lOOOi). W V = 10° ^25^"^ ~ ^°' " ^^■
(c) 2/ = 10 sin (I - .*) . (/) 2/ = 100 sin(5x + 4t).
2,r (?) y = 0-025 sin ^(,x + </3).
id) y = 50smy(a; - 3t). ■*
2. Write the equation of a progressive sinusoidal wave whose height
is 5 feet, length 40 feet and velocity 4 miles per hour.
3. Write the equation of a wave of wave length 10 meters, height 1
meter, and velocity of propagation 3.5 miles per hour. (Note: 1
mile = 1.609 kilometers.)
350 ELEMENTARY MATHEMATICAL ANALYSIS [§193
4. Sound waves of all wave lengths travel in still air at 70° F. with
a velocity of 1130 feet per second. Find the wave length of sound
waves of frequencies 256, 128, and 600 per second.
193. Stationary Waves. The form of a violin string during its
free vibration is sinusoidal, but the nodes, crests, troughs, etc.,
are stationary and not progressive as in the case of the waves
just discussed. Such waves are called stationary waves. The
water in a basin or even in a large pond or lake is also capable of
vibrating in this way. Fig. 144 may be used to illustrate the
stationary waves of this type, either of a musical string or of the
water surface of & lake, but in the case of a vibrating string, the
ends must be supposed to be fastened at the points 0 and N.
The shores of the lake may be taken at / and K-ot at I and H,
etc. As is well known, such bodies are capable of vibrating in
Fig. 144. — A stationary wave.
segments so that the number of nodes may be large. This
explains the "harmonics" of a vibrating violin string and the
various modes in which stationary waves may exist on a water
surface. A stationary wave on the surface of a lake or pond is
known as a seiche, and was first noted and studied on Lake
Geneva, Switzerland. The amplitudes of seiches are usually
small, and must be studied by means of recording instruments
so set up that the influence of progressive waves is eliminated.
The maximum seiche recorded on Lake Geneva was about 6 feet,
although the ordinary amplitude is only a few centimeters.
The equation of a stationary wave may be found by adding the
ordinates of a progressive wave
y = asin (hx — kt) (1)
§193] SIMPLE HARMONIC MOTION AND WAVES 351
traveling to the right (A; > 0), to the ordinates of a progressive
wave
y = asm {hx + /c<) (2)
traveling to the left.
Expanding the right members of (1) and (2) by the addition
formula for the sine and adding
y = 2a cos kt sin hx, (3)
or in terms of L and T, §188 (1) and §189 (1),
' y = 2a cos (^) sin {~:j ■ (4)
In Fig. 144, the origin is at 0 and the X-axis is the Line of nodes
ONX. If in equation (3) we look upon 2a cos kt as the vari-
able amplitude of the sinusoid
y = sin hx,
we note that the nodes, of the sinusoid remain stationary, but
that the amplitude 2a cos kt changes as time goes on. When
t = 0, the sine curve has amplitude 2a and wave length 2ir/h.
When t = ir/2k, or T/i, the sinusoid is reduced to the straight
line y = 0. When t = ir/k, or T/2, the curve is the sinusoid
y = — 2a sin hx
which has a trough where the initial form had a crest, or vice
versa.
Exercises
In the following exercises the height of the wave means the maxi-
mum rise above the line of nodes. When a seiche is uninodal, the
shores of the lake correspond to the points I and K, Fig. 144. When
a seiche is binodal, the points / and H are at the lake shore.
1. From the equation of a stationary wave in the form y =
2a sin %rx/L cos 2-wtlT, show that K, Fig. 144, is at its lowest depth
fori = r/2,,3r/2, 67/2, .
2. Henry observed a fifteen-hour uninodal seiche in Lake Erie,
which was 396 kilometers in length. Write the equation of the prin-
cipal or uninodal stationary wave if the amplitude of the seiche was
15 cm.
3. A small pond 111 meters in length was observed by Eridros to
have a uninodal seiche of period fourteen seconds. Write the equation
of the stationary wave if the ampUtude be o.
352 ELEMENTARY MATHEMATICAL ANALYSIS [§194
4. Forel reports that the uninodal longitudinal seiche of Lake
Geneva has a period of seventy-three minutes and that the binodal
seiche has a period of thirty-five and one-half minutes. The trans-
verse seiche has a period of ten minutes for the uninodal and five
minutes for the binodal. The longitudinal and transverse axes of the
lake are 45 miles and 5 miles respectively. Write the equation of
these different seiches.
5. A standing wave or uninodal seiche exists on Lake Mendota of
period twenty-two minutes. If the maximum height is 8 inches and
the distance .across the lake is 6 miles, write the equation of the seiche.
194.* Compound Harmonic Motion and Compound Waves.
The addition of two or more simple harmonic functions of frequencies
which are multiples of the frequency of a given first or fundamental
harmonic, gives rise to compound harmonic motion. Thus,
y = a sin fc< + & sin Zkt,
corresponds to the superposition of a S.H.M. of period 2ir/3fc and
amplitude 6 upon a fundamental S.H.M. of period 2ir/A; and amplitude
a. To compound motions of this type, there correspond compound
waves of various sorts, such as a fundamental sound wave with
overtones, or tidal waves in restricted bays or harbors. The graphs
of the curves
y = sin X + sin 2x
y = ainx + sin 3a
are easily constructed. They may be drawn by adding the ordinates
of the various sinusoids constructed on the same axis, as in Fig. 145.
To compound the curves, first draw the component curves, say y =
sin X and y = sin 3x of Kg. 145. Then use the edge of a piece of paper
divided proportionally to sin x (that is, like the scale OB, Fig. 145) and
use this as a scale by means of which the successive ordinates of a given
X may be added. For example, to locate the point on the composite
curve corresponding to the abscissa OD, Fig. 145, we must add DP
and DQ. Hence place vertically at P the lower end of the paper scale
just mentioned. The sixth scale division above P on this scale will
then locate the required point M of the composite scale.'
In Fig. 146 the curves:
y = sin X + sin (2x + 27rn/16)
y = sin 2x + sin (3x + 2)rre/16)
I Note that if the method described be used, there is really no need of drawing
the curve y = sin 3a:. If both curves are drawn, ordinates may conveniently be
added with bow dividers.
§194] SIMPLE HARMONIC MOTION AND WAVES 353
are shown for values of n = 0, 1, 2, . , IS in succession — that is,
for successive phase differences corresponding to one-sixteenth of the
wave length of the fundamental y = sin x.
Fig. 145. — The curves y = sin x,y= sin 3x and the compound curve
y = sin X + sin 3x. '^
Fifth
Fig. 146. — The curves (o) j/ = sin x + sin {2x + 27rn/16) and (b)
2/ = sin 2a; + sm (3x + 2Tn/16), for n = 0, 1, 2, . . 15. {From
Thomson and Tail.)
Wave forms compounded from the odd harmonics only are espe-
eially important, as alternating-current curves are of this type. See
Fig. 147.
23
354 ELEMENTARY MATHEMATICAL ANALYSIS [§196
196.* Harmonic Analysis. Fourier showed in 1822 in his "Ana-
lytical Theory of Heat" that a periodic single-valued function, say
y — f(x), under certain conditions of continuity, can be represented
by the sum of a series of sines and cosines of the multiple angles of the
form
y = ao + ai cos x + a^ cos 2x + Oa cos 3x + . . .
+ bi sin X +bi sin 2a; -|- 63 sin 3a; + . . .
This means, for example, that it is always possible to represent the
complex tidal wave in a harbor, by means of the sum of a number of
simple waves or harmonics. The term harmonic analysis is given to
the process of determining these sinusoidal components of a compound
periodic curve. In §194 we have performed the direct operation of
50
V
/
/
\
26
/
\
/
s
/
\
S
0
/
■ 1
) 1
2 1
1 1
6H
) !
2 2
4 2
i 2
i £
0 3
! 3
V
\
>
25
\
/.
V
/
60
\
/
~"
'
Fig. 147. — An alternating current curve.
present.
Only odd harmonics are
finding the compound curve when the component harmonics are given.
The inverse operation of finding the components when the compound
curve is given is much more difficult, and its discussion must be post-
poned to a later course.
196.* Connecting Rod Motion. If one end of a straight Une B be
required to move on a circle while the other end of the line A moves on
a straight Une passing through the center of the circle, the resulting
motion is Icnown as connecting rod motion. The connecting rod of a
steam engine has this motion, as the end attached to the crank travels
in a circle while the end attached to the cross-head travels in a
straight line. The motion of the end A, Fig. 148, of the connecting
rod is approximately S.H.M. The approximation is very close if the
§196] SIMPLE HARMONIC MOTION AND WAVES 355
connecting rod be very long in comparison with the diameter of the
circle.
A second approximation to the motion of the point A can be
obtained by introducing the second harmonic or octave of the funda-
mental. In Fig. 148, let the radius of the circle be a and the length
of the connecting rod be I. The length of the stroke M'N is 2a, and
the origin may conveniently be taken at the mid-point of the stroke,
0. When B is at E, A is at M and when B is at K, A is at A'^.
Then MH = NK = I and OC = I. Now
But
and
Hence
X = CA - CO = CA - I = CD + DA - I.
CD = a cos e
DA = Vl^ - BD' = Vl' - a^ sin^ e.
X = acose +1 Vl - (a^/l^) sin^ B - I
(1)
(2)
(3)
(4)
Fig. 148. — Connecting rod motion.
Approximating the radical by §113 (\/l — x = 1 — x/2) we obtain
^ , , / , a^ sin^ e\ , ,,,
X = acos 9 +1 il 2p — ) ~ '• (^)
Since sin^ 9 = (1 — cos 26) /2, we obtain
X = a cos 9+27 "^"^ ^^ ~ 47' ^^^
which is approximately true as long as I is much greater than a.
It is seen from the above result that the second approximation to
connecting rod motion contains as overtone the octave, or second
a*
harmonic, ^j cos 29, in addition to the first or fundamental harmonic
a cos 8.
356 ELEMENTARY MATHEMATICAL ANALYSIS [196 §
Exercises
1. Draw the curve corresponding to equation (5) above if o' = 1.15
inches, and Z = 3 inches.
2. The motion of a slide valve is given by an equation of the form
3/ = oi sin (9 + e) + 02 sin (28 + 90°).
Draw the curve if ai = 100, oj = 25, c = 40°, using 6 as the abscissa
in rectangular coordinates.
CHAPTER XII
COMPLEX NUMBERS
197. ScaJes of Numbers. To measure any magnitude, we
apply a unit of measure and then express the result in terms of
numbers. Thus, to measure the volume of the liquid in a cask
we may draw off the liquid, a measure full at a time, in a gallon
measure, and conclude, for example, that the number of gallons
is 125. In this case the number 12^ is taken from the arith-
metical scale of numbers, 0, 1, 2, 3, 4, . . If we desire to meas-
ure the height of a stake above the ground, we may apply a foot-
rule and say, for example, that the height in inches above the
ground is 12|, or, if the positive sign indicates height above the
ground, we may say that the height in inches is -|- 12J. In
this latter case the number -|- 12? has been selected from the
algebraic scale of numbers . . . — 4, — 3, — 2, — 1, 0, + 1,
+ 2, + 3, + 4, . .
The scale of numbers which must be used to express the value of a
magnitude depends entirely upon the nature of the magnitude. The
attempt to express certain magnitudes by means of numbers taken
from the algebraic scale may sometimes lead, as every student of
algebra knows, to meaningless absurdities. Thus a problem involving
the number of sheep in a pen, or the number of marbles in a box, or
the number of gallons in a cask, cannot lead to a negative result, for
the magnitudes just named are arithmetical quantities and their meas-
urement leads to a number taken from the arithmetical scale. The
absurdity that sometimes appears in results to problems concerning
these magnitudes is due to the fact that one attempts to apply the
notion of algebraic number to a magnitude that does not permit of it.
Science deals with a great many different kinds of magnitudes, the
measurement of some of which leads to arithmetical numbers while the
measurement of others leads to algebraic numbers; the remarkable
fact is that two different number scales serve adequately to express
magnitudes of so many different sorts.' The magnitudes of science
357
358 ELEMENTARY MATHEMATICAL ANALYSIS [§198
are so various in kind that one might reasonably expect that the
variety of number systems required in the mathematics of these
sciences would be very great.
The arithmetical scale is used when we enumerate the number of
gallons in a cask and say: 0, 1, 2, 3, . . . If we observe 3 gallons
in the cask, and then remove one, we note those remaining and say
tiDo; we may remove another gallon and say one, we may remove the
last gallon and say zero; but now the magnitude has come to an end.
The algebraic scale is used when we measure in inches the height of
a stake above the ground and say three. We may drive the stake
down an inch and say two; we may drive the stake another inch and
say one; we may drive the stake another inch and say zero, or
"level with the ground;" but, unUke the case of the gallons in the
cask, we need not stop but may drive the stake another inch and say
one below the ground, or, for brevity, minus one; and so on.
Many of the magnitudes considered in science are completely ex-
pressed by means of arithmetical numbers only; for example, such
magnitudes as density or specific gravity; temperature;^ electrical re-
sistance; quantity of energy; such as ergs, joules or foot-pounds;
power, such as horse power, kilowatts, etc. All of the magnitudes
just mentioned are scalar, as it is called; that is, they exist in one
sense only — ^not in one sense and also in the opposite sense, as do forces,
velocitiesj distances, as explained above. The arithmetical scale of
numbers is therefore ample for their expression.
The distraction, then, between an algebraic number and an arith-
metical number is the notion of sense which must always be associated
with any algebraic number. Thus an algebraic number not only
answers the question "how many" but also affirms the sense in which
that number is to be understood; thus the algebraic number -|- 12 J, if
arising in the measurement of angular magnitude, refers to an angular
magnitude of 12| units (degrees, or radians, etc.) taken in the sense
defined as positive rotation.
198. Algebraic Number Not the Most General Sort. Algebraic
numbers, although more general than arithmetical numbers, are
themselves quite restricted. For, each algebraic number corre-
sponds to a point of the algebraic scale. But for points not on the
scale there corresponds no algebraic number. That is, the alge-
braic scale is one-dimensional. It is thus seen that there is an
^ Temperature is an arithmetical quantity, since there is an absolute zero of
temperature. Temperature does not exist in two opposite senses, but in a single
sense.
§199] COMPLEX NUMBERS 359
opportunity of enlarging our conception of number if we can re-
move the restriction of one dimension — that is, if we can get out of
the line of the algebraic scale and set up a number system such that
one number of the system will correspond, for examiple, to each
point of a plane, and such that one point of the plane will corre-
spond to each number of the system. We will seek, therefore, an
extension or generalization of the number system of algebra that
will enable us to consider, along with the points of the algebraic
scale, those points which lie without it.
199. Numbers as Operators. The extension of the number
system mentioned in the last section may be facilitated by chang-
ing the conception usually associated with symbols of number.
The usual distinction in algebra is between symbols of number and
symbols of operation. Thus a symbol which may be looked upon as
answering the question "how many" is called a number, whUe a
symbol which tells us to do something is called a symbol of opera-
tion, or, simply, an operator. Thus in the expression -\/2) "v/ is
a symbol of operation and 2 is a number. A symbol of operation
may always be read as a verb in the imperative mood; thus we
may read -s/x: "Take the square root of x." Likewise log x,
and cos 9 may be read; "Find the logarithm of x," "Take the co-
sine of 8." In these expressions "log" and "cos" are symbols
of operation; they teU us to do something; they do not answer the
question "how many" or "how much" and hence are not num-
bers. Here we speak of -\/j log, cos, as operators ; we speak of x as
. the operand, or that which is operated upon.
It is interesting to note that any number may be regarded as a
symbol of operation; by doing so we very greatly enlarge some
original conceptions. Thus, 10 may be regarded not only as ten,
answering the question "how many," but it may quite as well be
regarded as denoting the operation of taking unity, or any other
operand that follows" it, ten times; to express this we may write
10-1, in which 10 may be called a tensor (that is, "stretcher"),
or a symbol of the operation of stretching a unit until the result
obtained is tenfold the size of the unit itself. In the same way
the symbol 2 may be looked upon as denoting the operation of
doubling unity, or of doubling any operand that follows it; like-
360 ELEMENTARY MATHEMATICAL ANALYSIS [§199
wise the tensor 3 may be looked upon as a trebler, 4 as a quadrupler,
etc.
With the usual understanding that any symbol of operation
operates upon that which follows it, we may write compound
operators like 2-2-3-1. Here 3 denotes a trebler and 31 denotes
that the unit is to be trebled, 2 denotes that this result is to be
doubled and the next 2 denotes that this result is to be doubled.
Thus representing the unit by a line running to the right, we have
the following representation of the operators :
The unit ->
3-1 -^^-^
2-31 > >
2-2-31 T > >
Notice the significance that should now be assigned to an expo-
nent attached to these (or other) symbols of operation. The
exponent means to repeat the operation designated by the operator;
that is, the operation designated by the base is to be performed,
and performed again on this result, and so on, the number of opera-
tions being denoted by the exponent. Thus W means to perform
the operation of repeating unity ten times (indicated by 10) and
then to perform the operation of repeating the result ten times,
that is, it means 10 (101). Also, 10' means 10[10(10-1)]. Like-
wise log^ 30 means log (log 30) which, if the base be 10, equals
log 1.4771, or finally 0.1694. An apparent exception- occurs in
the case of the trigonometric functions. The expression cos'j;
should mean, in this notation, cos (cos x), but because trigo-
nometry is historically so much older than the ideas here ex-
pressed, the expression cos'' x came to be used for (cos a;)', or
(cos x) X (cos x), but cos~^ 6 means arc cos 6, not 1/cos 9.
To be consistent with the notation of elementary mathematics,
the expression \/4, looked upon as a symbol of operation, must
denote an operation which must be performed twice in order
to be equivalent to the operation of quadrupling; that is, such
that (-\/4)^ = 4. Likewise i/i denotes an operation which
must be' performed three times in succession in order to be
equivalent to quadrupling. But we know that the operation
denoted by 2, if performed twice, is equivalent to quadrupling;
§200] COMPLEX NUMBERS 361
therefore \/4 = 2, etc. Just as 4^, 4', etc., may be called stronger
tensors than a single 4, so -s/i, Vi, etc. may be called weaker
tensors than the operator 4.
200. Reversor. The expression ( — 1), looked upon as a
symbol of operation, is not a tensor, as it leaves the size unchanged
of that upon which it operates. If this operator be applied to
any magnitude, it will change the sense in which the magnitude
is then taken to exactly the opposite sense. Thus, if 6 stands
for six hours after, then ( — 1)(6) stands for six hours before
a certain event, and ( — 1) is the sj'mbol of this operation of
reversing the sense of the magnitude. Also if 6 stands for a line
running six units to the right of a certain point, then ( — 1)(6)
stands for a line running six units to the left of that point; so
that ( — 1) is the symbol which denotes the operation of turning
the straight line through 180°. As 2, 3, 4, when looked upon as
symbols of operations, were called tensors, the operator ( — 1)
may conveniently be designated a reversor.
Exercises
Show graphically the effect of the operations indicated in each of
the following exercises. Take as the initial unit-operand a straight
line 1/2 inch long extending to the right of the zero or initial point.
Explaia each expression as consisting of the operand unity and
symbols of operation — ^tensors, reversors, etc., which operate upon
it, one after the other, in a definite order.
1. 2-3-1. 8. (Viy-i - 1)-1-
2. 3-3-1. 9. ( - l)s-22-31.
3. - 1-31. 10. 3-321.
4. 2'1. 11. ( - 1)'2-2«1.
6. VSI. 12. 3( - 1)V21.
6. (-v/2)^-l. 13. (\/2)-( - 1)"»-1.
7. -v/gVi-l. 14. Vl0-2-( - 1)1.
15. A tensor, if permitted to operate seven times in succession, will
just double the operand. Symbolize this tensor.
16. A tensor, if permitted to operate five times in succession, will
quadruple the operand. Symbohze this tensor.
362 ELEMENTARY MATHEMATICAL ANALYSIS [§201
201. Versors. The expression ■%/ — 1 cannot consistently,
with the meaning abeady assigned to \/ and ( — 1), be looked
upon as answering the question "how many," and therefore is not
a number in that sense; yet if we consider \/ — 1 as a symbol of
operation, it can be given a meaning consistent with the operators
already considered. For if 2 is the operator that doubles, and
\/2 is the operator that when used twice doubles, then since ( — 1)
is the operator that reverses, the expression \/ — ^ should be an
operator which, when used twice, reverses. So, as ( — 1) may
be defined as the symbol which operates to turn a straight line
through an angle of 180°, in a similar way we may define the
expression ■%/ — 1 as « symbol
which denotes the operation of
turning a straight line through
an angle of 90° in the positive
direction. The restriction of
positive rotation is inserted
to make the definition unique.
The symbols ( — 1) and
■y/ — 1 are not tensors. They
do not represent a stretching
or contracting of the operand.
Their effect is merely to turn
the operand to a new direc-
tion; hence these symbols
may be called versors, or
"turners."
202. The Operator V^^. In Fig. 149 let a be any line.
Then a operated upon by V - 1 (that is, V — 1 a) is a turned
anti-clockwise through 90°, which gives OB. Now, of course,
V— 1 can operate on V — 1 a just as well as on a. Then
V — 1 ( V — 1 a), or PC, is V — 1 g turned positively through
90°. Similarly, V - UV - 1(V - 1 a)] is V^I i.s/'^l a)
turned through 90°, etc.
As we are at liberty to consider two turns of 90° as equivalent
to one turn of 180°, therefore, \/ — 1 (V — 1 a) = ( — 1) o.
Now OD = ( - 1) OS, OD = i- 1) (-v/^T a); but also 0D =
B
(\Rfa
e
0 a J
C
9
J
'd
Fig. 149.-
-The integral powers of
§203] COMPLEX NUMBERS 363
V^ ( - a), therefore, ( - 1) V^ a = V"^ ( - «)• Thus
the student may show many like relations.
The operator •%/ — 1 is usually represented by the symbol i and
will generally be so represented in what follows.
Exercises
Interpret each of the following expressions as a symbol of operation:
1. 2, 3, 4, -1.
■2 3^23,4", (-1^ (-1)^
3. V2,VZ,V- 1, \/'2, \/- 1.
Select a convenient unit and construct each of the following expres-
sions geometrically, explaining the meaning of each operator:
4. 2-3-5-1. 7. (-1)''V^^-1.
6. 2=-(-l)-l. 8. 2'-(-l)^-(\/- l)"-!.
6. 3V-1-21. 9. 3V - 1(-1)V -11.
203. Complex Numbers. An expression of the form a + hi
is cdlled a complex number, since it contains a term taken from
each of the following scales, so th.at the unit is not single but
double or complex:
- 3, ■- 2, - 1, 0, + 1, + 2, + 3,
. - 3i, - 2i, - i, 0, + I, + 2i, + 3t,
Any number belonging to the first scale is called a real nimiber,
any number belonging to the second scale is called a pure
imaginary.
It is important to note that the only element common to the two
series in this complex scale is 0.
The explanation of the meaning of the symbol (a + hi) will
be given in the following section. It will be shown in subsequent
theorems that any expression made up of the sum, product,
power or quotient of complex numbers may be put in the form
a + hi, in which both a and 6 are re&l.
204. Meaning of a Complex Nimiber. Any real number, or
any expression containing only real numbers, may be consid-
ered as locating a point in a line.
Thus, suppose we wish to draw the expression 2 + 5. Let 0 be
364 ELEMENTARY MATHEMATICAL ANALYSIS [§204
the zero point and OX the positive direction. Lay off OA = 2 in
the direction OX and at A lay off AB = 5 in the direction OX.
Then the path OA + AB is the geometrical representation of
2+5.
0 A B X
Any complex number may be taken as the representation of the
position of a point in a plane. For, suppose c + di is the complex
number. Let 0, Fig. 150, be the zero point and OX the positive
direction. Lay off OA = + c in the direction OX and at A erect
di in the direction OY, in-
stead of in the direction OX
as in the last example. It
is agreed to consider the
step to the right, OA,
followed by the step up-
ward, AP, as the meaning
of the complex number c +
di^ Either the broken 'path
OA + AP or the direct -path
OP may he taken as the repre-
smtation of c + di, and either
path constitutes the definition
of the sum of c and di.
— di, and — c + di may be
Fig. 150. — The geometrical con-
struction of a complex number,
c + di.
di.
In the same manner c
constructed.
The meaning of some of the laws of algebra as applied to imagi-
naries may now be illustrated. Let us construct c + di + a + hi.
The first two terms, c + di, give OA + AB, locating B (Fig.
151). The next two terms, a + hi, give BC + CP, locating P.
Hence the entire expression locates the point P with reference to
0. Now if the original expression be changed in any manner
allowed by the laws of algebra, the result is merely a different path
to the same point. Thus:
c + a +di + hiis the path OA, AD, DC, CP
{c+a)+ {d + h)i is the path OD, DP
a + di+ c + 6i is the path OE, EH, HC, CP
a+di + hi + c is the path OE, EH, HF, FP, etc.
§205]
COMPLEX NUMBERS
365
The student should consider other cases. Is there any method
of locating P with the same four elements, which the figure does
not illustrate?
205. Laws. It can be shown by simple geometrical construc-
tion that the operator i, as defined above, obeys the ordinary
laws of algebra. We can then apply all of the elementary laws of
algebra to the symbol i and work with it just as we do with any
other letter. The following are illustrations of each law:
r a
c
y
^ ^
^
.
1
F
G
f
•*
^-N
•a
I
H
B
a
c
ts
•■s
^
0
E
^
,
.
A
D
Fig. 151. — Illustration of the application of the laws of -algebra and
the expression c + di + a + bi.
CoMMUTATrvE Law:
c-\-di-\-a + bi = c + a + di-\-bi = di-{-c + bi + a, etc.
ai = ia, iai = iia = aii, etc.
Thus the equation lO-s/ — 1
/ - llO.or better, lOV - 1-1
\/ — l-lOl may be said to mean that the result of performing
the operation of turning unity through 90° and performing upon ,
the result the operation of taking it ten times, is the same as the
result of performing the operation of taking unity ten times and
performing upon this result the operation of turning through 90°.
AssociATivB Law:
(c + di) + (a + bi) = c + {di + a) + bi, etc.
{ab)i = a{bi) = abi, etc.
DiSTEiBUTrvE Law:
(a + b)i = ai +
etc.
366 ELEMENTARY MATHEMATICAL ANALYSIS [§206
The expression -\/ — a, where a is any number of the arith-
metical scale, is defined as equivalent to \/ — l-o;that is, y/ — a
- i\fa. For example, V — 4 = 2i, V —3 = i'\/^, etc. In
what foUows it is presupposed that the student will reduce expressions
of the form -y/ — ato the form i s/a before performing algebraic op-
erations. From this it follows that y/ — a-^/ — b = — y/cA and
not Vobl
The relation -\/ — 4 = 2-\/ — 1 may be interpreted as follows :
( — 4) is the operator that quadruples and reverses; then •%/ — 4
is an operator which used twice quadruples and reverses. But
2-%/ — 1 is an operator such that two such operators quadruple
and reverse. That is, V — 4 = 2\/ — 1.
206. Powers of i. We shall now interpret the powers of i by
means of the new significance of an exponent and by the commu-
tative, associative and other laws. First:
i° or i° 1 = + 1 i^ = iH = i
^ i' .or i^ 1 = i %'• = iH = — \
j2 = _ ]^ j7 ~= m = — i
i^ = iH = — i i' = m = + 1
i* = m^ = + 1 etc. etc.
Whence it is seen that all even powers of i are either + 1 or — 1,
and all odd powers are either i or — i. The student may reconcile
this with Fig. 149. The zero power of i must be unity, for the
exponent zero can only mean that the operation denoted by the
symbol of operation is not to be performed at all; that is, unity is to
be left unchanged; thus 10° or 10»-1 = 1, 2" = 1, log" x = x,
sin" X = X, etc.
Exercises
Select as unit a distance 1/2 inch in length extending to the right
and represent graphically each of the following expressions:
1. i + 2i' + 3i' + 4i* -f .
2. t + i« + i* + i« + i' + .
3. i + i* + e + i^ + i' + i'^ + .
4. i(i + i< + i' + i* + i' + ii2 + . . ).
5. i + i« -f- 1' + 2i^ + i* + t" + i' + 3i» + . . .
§^07] ," COMPLEX NUMBERS 367
207. Conjugate Complex Numbers. Two complex numbers
are said to be conjugate if they differ ohiy in the sign of the term
containing \/ — 1." Such are x + iy and x — iy.
Conjugate imaginaries have a real sum and a real product.
For {x + yi) + {x — yi) = x + yi + x — yi,
=. X + X + yi — yi = 2x.
Likewise, applying the ordinary rules of algebra,
{x + yi) (x — yi) = x^ — yH'' = a;^ + j/^
It is well to note that the product of two conjugate complex
numbers is always positive and is the sum of two squares.
This fact is very important and will be used frequently. Thus
(3 - 4i)(3 + 4.1) = 3^'+ 42 = 25; (1 + i){l - i) = 2;
(cos d + i sin 9) (cos d — i sin 6) = cos" 6 + sin" 0 = 1; etc.
208. The sum, product, or quotient of two complex numbers is,
in general, a complex number of the typical form a + bi.
Let the two complex numbers be a; + yi and u + vi.
(1) Their sum is (x + yi) + (u + vi)
= (x + u) + {y + v)i
by the laws of algebra. This last expression is in the form a + bi.
(2) Their product is {x + yi) (u + vi)
= x{u + vi)+ yi{u + vi)
= xu + xvi + yui + yvi'
= {xu — yv) + {xv + yu)i
by the laws of algebra. This last expression is in the form a + bi
(3) Their quotient is
X + yi _ (x + yi){u — vi)
u + vi (m + vi)(u — vi)
By the preceding, the numerator is of the form a' + b'i. By
§207, the denominator equals m" + «". Then the quotient equals
a' + b'i a' b' .
u^ + v^ m" + w" ' m" + »2
by distributive law. This last expression is of the form a + bi.
368 ELEMENTARY MATHEMATICAL ANALYSIS
Exercises
Reduce the following expressions to the typiqal form a + bi; the
student must change every imaginary of the form -y/ — o to the form
1. V - 25 + V~^^ + V^^i2i - V^'ei - 6i.
2. (2V~^^ + 3v'^)(4\/"-^3 - 5V^^).
3. (x - [2 +3i]){x - [2 -3i]).
4. (-5 + 12V^T)^. 6. (vr+i)(-v/r^).
5. (3 - 4V^.)'. 7. (Ve"- V^~^)'.
a 1
8. , . 12.
2 1 - i^
13.;
"■ S+V -2 • (1 - 0'-
10. , 'V _ 14. l^.^^A
11. 1 +V 15. (2 + sV^^n.^
i-i" 2 + v^n" ■
-„ o + a;i a — xi
lb. ^ j ;•
a — XI a + x%
209. If a complex number is equal to zero, the imaginary and
real "parts are separately equal to zero.
Suppose X + y \/ — 1 = 0,
X and y being real numbers.
Then x = — y V — 1.
Now it is absurd or impossible that a real number should equal
an imaginary, except they each be zero, since the real and imagi-
nary scales are at right angles to each other and intersect only at
the point zero.
Therefore x = 0 and y = 0.
If two complex numbers are equal, then the real parts and the
imaginary parts must be respectively equal.
For if X + yi = u + vi
then (x -«) + (?/ - v)i = 0.
§2101
COMPLEX NUMBERS
369
Whence, by the above theorem,
That is,
X — u = 0 and y — v= 0.
a; = M and y = v.
210. Modulus. Let the complex number x + yihe constructed,
as in Fig. 152, in which OA = x and AP = yi. Draw the line
OP, and let the angle AOP be called 0.
The numerical length of OP is called the modulus of the complex
number x + yi. It is algebraically represented by -y/x^ + y^,
or by the symbol \x + yi\. Thus, mod (3 + 4*) = V9 + 16 = 5.
The student can easily see that two conjugate complex numbers
have the same modulus.
If 2/ = 0, the mod (x + yi) = \/^= \x\, where the vertical
lines indicate that merely the numerical, or absolute, value of
X is called for. Thus the
modulus of any real number
is the same as what is called
the numerical, or absolute
value, of the number. Thus
mod (— 5) = 5.
211. Amplitude. In Fig.
152 the angle AOP or 6 is
called the argument, or ampli-
tude, or simply the angle, of
the complex number x + yi.
Putting r = \^x^ + y^ - mod {x +
y
Fig.
152. — Modulus and amplitude
of a complex number.
yi) = ;x +iy\, we have
sin 6 =
and
cos 6
X
r
Therefore,
.•B + ?/i = r cos 0 + ir sin Q = r(cos 9 + i sin 9).
(1)
We have expressed the complex number x + yi in terms of its
modulus and amplitude. The last member of (1) is called the
polar fonn of the complex number {x + iy).
To put 3 — 4i in this form, we have
mod (3 - 4i) = \/9 + 16 = 5; sin 5 = ^ = - |; cos S = - = f
^ r 5 r 5
24
370 ELEMENTARY MATHEMATICAL ANALYSIS [§212
Therefore,
The amplitude d is tan-' ( ~ o ) i ^^^ is in the fourth quadrant.
Why?
It is well to plot the complex number in order to be sure of the
amplitude 6. It avoids confusion to use positive angles in all
cases. For example, to change 3 — \/3 i to the polar form, plot
the point (3, — \/3) and find from the triangle that r = 2 \/3 and
9 = 330°. Hence
3 - VS i = 2V3(cos 330° + i sin 330°). [<
The ampUtude of all positive numbers is 0, and of all negative
numbers is 180°. The unit expressed in terms of its modulus and
amplitude is evidently l(cos 0 + i sin 0).
212. Vector. The point P, Fig. 152, located by OA + AP, or
X + yi, may also be considered as located by the line or radius
vector OP; that is, by a line starting at 0, of length r and making
an angle 6 with the direction OX. A directed line, as we are now
considering OP, is called a vector. When thus considered, the two
parts of the compound operator
r (cos 5 + i sin 6) (1)
receive the following interpretation : The operator (cos 6 + ism 6),
which depends upon B alone, turns the unit Ijdng along OX
through an angle 6, and may therefore be looked upon as a versor
of rotative power 6. The versor (cos 6 -\- i sin 6) is often abbre-
viated by the convenient symbol cWd. The operator r is a tensor,
which stretches the turned unit in the ratio 1 : r. The result of
these two operations is that the point P is locaited r units from 0
in a direction making the angle 6 with OX.
Thus, the operator (cos ^ + i sin 6) is simply a more general
operator than i, but of the same kind. The operator i turns a
unit through a right angle and the operator (cos 0 -\- i sin 6) turns
a unit through an angle B. If 6 be put equal to 90°, cos 6-\-i sin 6
reduces to i.
§213] COMPLEX NUMBERS ' 371
For d = 0, cos 6 + i sin $ reduces to 1
6 = 90°, cos d + i sin d reduces to i
6 = 180°, cos 6 -\- i sin 6 reduces to — 1
6 = 270°, cos 6 + isin 6 reduces to — i
Since 3 — 4i = 5(f — fi), the point located by 3 — 4i may be
reached by turning the unit vector through an angle 8 =
sin~'(— 4/5) = COS"' 3/5 and stretching the result in the ratio 1 :5.
// a complex number vanishes, its modulus vanishes; and con-
versely, if the modulus vanishes, the complex number vanishes.
li X + yi = 0, then x = 0 and y = 0, hy §210. Therefore,
Vx^ + t/2 = 0. Also, if Va;^ + y^ = 0, then x^ + y^ = 0, and
since x and y are real, neither x' nor y^ is negative, and so their
sum is not zero unless each be zero.
// two complex numbers are equal, their moduli are equal, but if
two moduli are equal, the complex numbers are not necessarily equal.
li X -i- yi = u + vi, then x = u and y = vhy §210.
Therefore, V^^+^ = Vu^ + vK
But if ■y/x'^ + y'^ = y/u^ + v^, obviously x"^ need not equal
u^ nor y"^ = v'.
213. Sum of Complex Numbers. Let a given complex number
locate the point A, Fig. 153, and let a second complex number
locate the point B. Then if the first of the complex numbers be
represented by the radius vector OA, and if the second complex
number be represented by the radius vector OB, the sum of the
two complex numbers will be represented by the diagonal OC of
the parallelogram constructed on the lines OA and OB. This law
of addition is the well-known .law of addition of vectors used in
physics when the resultant of two forces or the resultant of two
velocities, two accelerations, or two directed magnitudes of any
kind, is to be found.
The proof that the sum of the two complex numbers is repre-
sented by the diagonal OC is very simple. Let the graph of the
first complex number be ODi + DiA and let that of the second be
OD2 -f- DiB. To add these, at the point A construct AE = ODi
and EC = D^B. Then the sum of the two complex numbers is
geometrically represented by OJ)^ + BiA + AE + EC, or by the
372 ELEMENTARY MATHEMATICAL ANALYSIS [§214
radius vector OC which joins the end points. Since, by construc-
tion, the triangle AEC is equal to the triangle OD^B, AC must be
equal and parallel to OB, and the figure OACB is a parallelogram.
OC, which represents the required sum, is the diagonal of this
parallelogram, which we were required to prove.
Di Di Di
Fig. 153. — Sum of two complex numbers.
Exercises
Mnd algebraically the sum of the following complex numbers, and
construct the same by means of the law of addition of vectors.
1. (1 + 2i) + (3 + 4i). 4. (3 - 4i) - (3 + 4i).
2. (1 + i) + (2 + i). 5. (-2 + i) + (0 - ti).
3. (1 - i) + (1 + 2i). 6. (- 1 + i) + (3 + i) + (2 + 2i).
7. (2 - i) + (- 2 + i) + (1 + i)-
8. Find the modulus and ampHtude ^in degrees and minutes) of
2(cos 30° .+ i sin 30°) + (cos 45° + i sin 45°).
9. By the parallelogram of vectors, show that the sum of two con-
jugate complex numbers is real.
10. If ij be the sum of the complex numbers Zi'= xi -|- iyi, Z8 =
Xi -\-iyi, «a = Sa + Vii, etc., show that —R, zi, zj, 23, . . . form the
sides of a closed polygon.
214. Product of Complex Nximbers. The product, of two or
more complex ^umbers is a complex number whose modulus is'>the
§214]
COMPLEX NUMBERS
373
product of the moduli and whose amplitude is the sum of the ampli-
tudes of the com,plex numbers.
Let the complex numbers be
^1 = xi + y-ii = ri (cos Q\ + i sin 6i)
Z2 = a;2 + 2/21 == rj (cos 02 + i sin ^2), etc.
By actual multiplication, we get
«i22 = rir-2 [(cos 01 cos 02 — sin 0i sin 02) +
(sin 01 cos 02 + cos 0i sin di)i\ = rir^ [cos (0i + 02) + i sin (0i + di)]
Whence it is seen that rir2 is the modulus of the product -and
(01 + 02) is the amplitude.
(2 + 2»)(v7+i)
Fig. 154. — Product of two pomplex numbers.
The above theorem is illustrated by Fig. 154. If the two given
complex numbers be represented by their vectors OPi and OPt, their
product will be represented by the vector OP3 whose direction angle
is the sum of the amplitudes of the two given factors, and whose
length OP3 is the product of the lengths OPi and OP2.
The figure represents the product (2 + 2i) {y/z + i). Expressed in
terms of modulus and ampUtude these may be written,
-s/3'+ i = 2(cos 30° + i sin 30°)
2 + 21 = 2v^(cos 45° + i sin 45°)
374 ELEMENTARY MATHEMATICAL ANALYSIS [§215
Hence, ri = 2, rj = 2\/2, Bi = 30°, 6, = 45°
Therefore (2 + 2i)(V3 + i) = 4v'2 (cos 75° + i sin 75°)
Exercises
Find the moduli and amplitudes of the following products, and
construct the factors and products graphically. Take a positi-i/e angle
for the amplitude in every case.
1. (1 + \/3t)(2-\/3 + 2i). 4. (1 + iy. _
2. (2 + W3i){2 + 2i). 5. (2 - 2v'3i)(\/3 + Si).
3. (V3 + 3i)(2 - 2i). 6. (1 - iy.
7. (1 + i)\l - i)K
8. 2 (cos 15° + i sin 15°) X 3 (cos 25° + i sin 25°).
Find numerical result by use of slide rule or trigonometric tables.
9. 2(cos 10° + i sin 10°) X (1/3) (cos 12° + i sin 12°) X
6(co3 8° +isin8°).
10. Find the value of 4-\/2 (cos 75° +isin75°) + (Vs + i).
216. Quotient of Two Complex Numbers. The quotient of
two complex numbers is a complex number whose modulus is the
quotient of the moduli and whose amplitude is the difference of the
amplitudes of the two complex numbers. Let the complex numbers
be
Si = Xi + yii = ri(cos Q\-\- i sin 9i)
Zi = Xi + yii = rjCcos di + i sin 62).
We have
zi _ ri(cos Bi + i sin gi)(cos dj — i sin 6i)
02 raCcos 02 + i sin S2)(cos 82 — i sin 0i)
^ ri[oos {di - 62) +i sin (9i - gg)]
r2 (cos'' ^2 + sin^ 62)
= -[cos (^1 -■^2) + i sin (^i - 62)]-
r2
Whence it is seen that — is the modulus of the quotient and
(.01 — 02) is the ampUtude.
In Fig. 155, the complex number represented by the vector OPi
when divided by the complex number represented by OP2 yields
the result represented by OP3, whose length ri/rais found by dividing
the length of OPi by the length of OP2, and whose, direction angle
§216]
COMPLEX NUMBERS
376
is the difference (ffi — 61) of the amplitudes of OPi and OPi.
The figure is drawn to scale for the case:
5 (cos 60° + i sin 60°)
2 (cos 20° + i sin 20°)
= (2.5) (cos 40° + i sin 40°)
Fig. 155. — Quotient of two complex numbers.
Exercises
Find the quotient and graph the results in each of the following
exercises. Always take ampUtudes as positive angles and if 9j > 61,
take 9i + 360° instead of 9i.
1. (1 + \/3i) -^ (V2 + V2i). 3. (SVS -3i) -i- ( - 1 + \/3i).
2. (i + iVSi) -^ (^2 - V2i). 4. (1 - VSi) -h i.
6. 2(oos 36° + i sin 36°) -r- 5(cos 4° + i sin 4°;.
6. 1.2(cos 48° + i sin 48°) h- [2(cos 15° + i sin 15°)
3(cos 9° + i sin 9°;].
„ [4 + (4/3)-v/3i] (2 + 2Vdi)
8 + 8i
8. Express in terms of a, b, e, d, the ampUtude of (a + bi) +
(c + di). I
216. De Moivre's Theorem. As a special case of §214 consider
the expression
(cos d + i sin S)»
where n is a positive number.
376 ELEMENTARY MATHEMATICAL ANALYSIS [§216
This being the product of n factors like (cos e + i sin e), we
write, by means of §214,
(cos S + i sin 6) (cos 0 + i sin S)
= [cos(0+e+ . )+isin((9+e+ ...)],
or
(cos 5 + i sin 6)" = (cos nd + i sin nd), (1)
which relation is known as De Moivre's theorem.
De Moivre's theorem holds for fractional values of n. For, first
consider the expression
(cos e + i sin e)^^\
where the power 1/t of aos d -\- i sin B is, by definition, an
operator such that the <th power of the expression equals
cos 0 + i sin B.
a
Put B = t(i>, SO that 4> — ~t
Then (cos B + i sin 9)'/' = (cos tij) + i sin tij))^'^
= [(cos ^ + i sin </.)']i'" by (1)
= cos <j> + i sin <p
= cos 1 + I sin 1 • (2)
Next consider the case in which n = j. We know
(cos B + i sin BY'' = [(cos fl + i sin B)')]^''
= (cos sB+ i sin sfl)!/' by (1)
= cosy +i sin y by (2). (3)
Likewise the theorem may be proved for negative values of n.
Illustkation 1. Find (3 + i \/3)*.
Write 3 + i VS = 2 V3(cos 30" + i sin 30°).
Then, by De Moivre's theorem,
(3 + i VS)* = 144(cos 120° + i sin_120°)
= 144( - l/2+_^V3i)
= -72 + 72-v/3i
§217] COMPLEX NUMBERS 377
Illustration 2. Knd (2 + 2i)".
Write 2 + 2i in the form
2 + 2i = 2 \/2_(i V2 + i V2i)
(2 + 2i)'i = (2 V2)"(cos 45° + i sin 45°)"
= (2 \/2)"(cos 495° + i sin 495°)
= (2 •v/2)"(ooa 135° + i sin 135°)
= (2V'2)"( - ^ ^2 +i\/2i)
= 2i« ( - 1 + i).
Exercises
Evaluate the following by De Moivre's theorem, using trigonomet-
ric table or slide rule when necessary.
1. (8 +8\/3t)".V 6. [1/2 + (l/2)V3i]*.
2. •>y27 (cos 76° - i sin 75°). 7. (1 + i)'.
3. -^125i. 8. (- 2 + 2i)^.
4. [cos 9° + i sin 9°]". 9. [(1/2)V3 - (l/2)i]'.
5. (S+VSi)'-
-10. Find value of (-1 + V - 3)= + (-1 - V - 3)' by De
Moivre's theorem.
11. Find the value of x^ - 2z + 2 for x = 1 + i.
12. If ii = - 1/2 + (1/2)^^^ and J2 = - 1/2 - (1/2) V - 3,
showthatjV = l.jV = l,ji' =J2,h^ =Ju3i^" =jV = l,jV"'^'=ii.
217. The Roots of Unity. Unity may be written
1 = cos 0 + i sin 0
1 = cos 2ir + i sin 2ir
1 = cos 47r + i sin Air
1 = cos 6t + i sin 6ir
and so on. By De Moivre's theorem the cube root of any of these
is taken by dividing the amplitudes by 3. Therefore, from the
above expressions in turn, there results
Vi = cos 0 + i sin 0 = 1
•^1 = cos (27r/3) + i sin (27r/3) = cos 120° + i sin 120°
= -l/2 + i(l/2) V3
Vl = cos (4ir/3) + i sin (47r/3) = cos 240° + i sin 240°
_ = - 1/2 - i(l/2) V3
Vl = cos 67r/3 + i sin Qir/S = same as first, etc.
378 ELEMENTARY MATHEMATICAL ANALYSIS [§217
Therefore there are three cube roots of unity. Since these are the
roots of the equation a;' — 1 = 0, they might have been found by
factoring, thus
x> - 1 = (x - 1) ix^ + x+ 1)
= ix-l){x + 1/2 + i Vsi) (^ + 1/2-4 V3i)
The three roots of unity divide the angular space about the point
0 into three equal angles, as shown in Fig. 156. In the same
way, it can be shown that there are four fourth roots, five fifth
roots, etc., of unity and that the vectors representing them have
modulus 1 and amplitudes that divide equally the space about 0.
B
\
~^
1 S
f ^
o /\
D
-ii\ /
y'
V 1
r
j^ \
c
^
-^-^
Fig. 156. — The cube roots of unity.
IlliTJStbatign 1. Find Vvl+3i
Write \/3 + 3i in the form
VS + 3i = 2-\/3(cos 60° + i sin 60°)
Hence, by De* Moivre's theorem,
WZ + 3i)^ = y/Vi (cos 30° + i sm 30°)
= -v/T2[(l/2)^V3 + (l/2)i]
= (1/2)1^108+ (1/2)^^12 i
A second root can be found by writing
VS + 3i = 2V'3 [cos (60° + 360°) - i ein (60° + 360°)]
|218] COMPLEX NUMBERS 379
siace adding a multiple of 360° to the amplitude does not change
the value of the sin and cosine. In applying De Moivre's theorem,
there results
(VS + 3i)>^ = ■>yi2 (cos 210° +i sin 210°) '
= V^12 [ - (l/2)v'3 - (l/2)i]
Illttstration 2. Find the cube root of — \/2 + \/2 i.
We write:
- \/2 + i V2 = 2 (cos 135° + i cos 135°)
= 2[cos (135° + Ji360°) + i cos (135° + 7i360°)].
in which n is any integer. Hence
{-\/2+i V2)^ = \/2 [cos (45° + ?il20°) + i sin (45° + nl20°)]^
= ^ (cos 45° + i sin 45°) for n = 0
= -^2 (cos 165° + i sin 165°) forra = 1
= -^2 (cos 285° + i sin 285°) for n = 2.
These are the three cube roots of the given complex number. For
« = 3 the first root is obtained a second time.
Exercises
Find all the indicated roots of the following:
1. (8 + SVsi)^. 6. (2 + 2i)^.
2. i^27(cos 75° + i sin 75°). 6. 32^.
3. -^iMi. 7. V/5I2.
4. ( - 2 + 2i)^.
8. Find to four places one of the imaginary 7th roots of + 1.
Note: Cos 51° 25.7' + i sin 51° 25.7' = 0.6235 + 0.7818i.
218.* Irrational Numbers. A rational ntunber is a number that
can be expressed as the quotient of two integers. All other real
numbers are irrational. Thus V^j \/5j V^?, ir, e, are irrational
numbers. An irrational number is always intermediate in value
to two rational numbers which differ from each other by a number
as small as we please. Thus
1.414, < •v/2 < 1.415
1.4142 < \/2 < 1.4143
1.41421 < V2 < 1.41422, etc.
380 ELEMENTARY MATHEMATICAL ANALYSIS [§218
It is easy to prove that \/2 cannot be expressed as the quotient
of two integers For, if possible, let
V2=l, (1)
where a and 6 are integers and r is in its lowest terms. Squaring
the members of (1) we have
2 =^. (2)
This cannot be true, since 2 is an integer and a and 6 are prime
to each other.
An irrational number, when expressed in the decimal scale, is
never a repeating decimal. For, if the irratiqn,al number could be
expressed in that manner, the repeating decimal could be evalu-
ated by §120 in the fractional form ^ _ ' which, by definition
of an irrational number, is impossible. On the contrary, every
rational number when expressed in the decimal scale is a repeating
decimal. Thus 1/3 = 0.33 . . and 1/4 = 0.25000. . .
The proof that ir and e are irrational numbers is not given in
this book.^
The student should not get the idea that because irrational num-
bers are usually approximated by decimal fractions, that the
irrational number itself is not exact. This can be illustrated by
the graphical construction of ■\/2. Locate the point P whose
coordinates are (1, 1). Call the abscissa OD and the ordinate DP.
Then OP = y/% OZ) = 1, and DP = 1. It is obvious that the
hypotenuse OP must be considered just as exact or definite as the
legs OB and DP The notion that irrational numbers are inexact
must be avoided.
The process of counting objects can be carried out by use of the
primitive scale of numbers 0, 1, 2, 3, 4, . . . The other numbers
made use of in mathematics, namely,
(1) positive and negative numbers
(2) integral and fractional numbers
(3) rational and irrational numbers
(4) real and imaginary numbers,
^ See Monographs on Modern Mathematics, edited by J. W. A. Young.
§219] COMPLEX NUMBERS 381
may be looked upon as classes of numbers that permit the opera-
tions siibtraction, division and evolution, to be carried out under all
circumstances. Thus, in the history of algebra it was found that
in order to carry out subtraction under all circumstances, negative
numbers were required; to carry out division under all circum-
stances, fractions, were required; to carry out evolution of arith-
metical numbers under all circumstances, irrational numbers
were required; finally to carry out evolution of algebraic numbers
under all circumstances, imaginaries were required. It will be
found that it wiE not be necessary to introduce any additional
form of number into algebra; that is, the most general number
required is a number of the form a 4- 6i, where a and 6 are positive
or negative, integral or fractional, rational or irrational. This is
the most general number that satisfies the following conditions:
(a) The possibility of performing the operations of algebra and
the inverse operations under all circumstances.
(6) The conservation or permanence of the fundamental laws of
algebra: namely, the commutative, associative, distributive, and
index laws.
Further extension of the number system beyond that of complex
numbers leads to operators which do not obey the commutative
law in multiplication; that is, in which the value of a product is
dependent upon the order of the factors, and in which a product
does not necessarily vanish when one factor is zero. Numbers of
this kind the student may later study in the introduction to the
study of electromagnetic theory under the head of "Vector ^
Analysis" or in the subject of "Quaternions.'' Such numbers or
operators do not belong to the domain of numbers we are now
studying.
219.* Simple Periodic Variation Represented by a Complex Num-
ber. Fluctuating magnitudes exist that follow the law of S.H.M.
although, strictly speaking, such magnitudes can be said to be "sim-
ple harmonic motions" in only a figurative sense. For example we
may think of the fluctuations of the voltage or amperage in an alter-
nating current as following such a law. Thus if E represent the
electromotive force or pressure of the alternating current, then the
fluctuations are expressed by
E = Eo sin oit
382 ELEMENTARY MATHEMATICAL ANALYSIS [§219
or by
E = Eo sin 2irft,
where/is the frequency of the fluctuation. Instead of S.H.M. such a
variable is naiore accurately called a sinusoidal varying magnitude,
although for brevity we shall often call it S.H.M. The graph in rec-
tangular coordinates of such a periodic function is often called a
"wave," although this term should, in exact language, be reserved for
a moving periodic curve, such a.sy = a sin (hx — kt).
If the polar representation
p = a sin (ot — <i) (1)
of the sinusoidal varying magnitude be used, then the graph of (1)
is a circle of diameter a inclined the angular amount uU to the left of
the axis OY, as is seen at once by calUng cot = 9 and aU = a in the
equation of the circle p = o sin (9 — a). The circle can be drawn
when the length and direction of its diameter are known; that is, the
circle is completely specified when a and the direction of a (told by
a) are given. Therefore the simple harmonic motion is completely
symbolized by a vector OA of length a drawn from the origin in the
direction given by the angle uti. The direction angle of the vector OA is
a + 2> or ah + g-
The circle on the vector OA is located or characterized equally
well if the rectangular coordinates (c, d) of the end of the diameter
of the circle be given. But the complex number c + diis represented
by a vector which coincides with the diameter o of this circle. Hence
we may represent the circle by the complex niunber c + di. Its
modulus is a = ■\/c' + d'' and its amplitude is a + s. Therefore if in
(1) we take o = \/c* + d-, at, = a and the variable angle at = 9, we
can completely describe the S.H.M. by the complex number c + di.
In the theory of alternating currents the sinusoidal varying current or
voltage can conveniently be represented by a complex number, and
that method of representing such magnitudes is in common use.
One of the advantages of representing S.H.M. by a vector or by a
complex number is the fact that two or more such motions of like
periods may then be compounded by the law of addition of vectors.
This method of find^g the resultant of two sinusoidal varying mag-
nitudes of like periods possesses remarkable utihty and simplicity.
To summarize, we may say:
(o) A siniisoidal varying magnitude is represented graphically in
§219]
COMPLEX NUMBERS
383
polar coordinates by a vector, which by its length denotes the amplitude
and by its direction angle with respect to OY denotes the epoch angle.
(6) Sinusoidal varying magnitudes of like periods may be compounded
or resolved graphically by the law of parallelogram of vectors.
If two sinusoidal varying magnitudes of like periods are in quad-
rature (that is, if their epoch angles differ by 90°), their relation,
neglecting their epochs, can be completely expressed by a single com-
plex number. Thus let two S.H.M. in quadrature
and
E„ = 113 sin a{t - h)
Ec = 40 cos a{t - ti)
(2)
(3)
Fig. 157. — Composition of two S.H.M. in quadrature by law of
addition of vectors.
be represented by the circles and by the vectors marked OEo and
0B„, Fig. 157. Call the resultant of these Ei. Then
Ei = 113 sin fc)(i - «i) + 40 cos u(« - <i) (4)
= -\/402 -t- 113^ sin w{t - ti)
= 120 sin w{t - ti), (5)
where wij is measured as shown in Fig. 157. Instead of representing
(2) and (3) in the polar diagram by 0E„ and OEe and their resultant
DX
384 ELEMENTARY MATHEMATICAL ANALYSIS [§220
by OEi, we may represent (2), (3), and (4) in the complex number
diagram, Fig. 158, by £?„, lEc, and Ec + iEc, respectively. Since the
modulus and amplitude of £„ + iEc are y/Eo' + &* and a, respec-
tively, and since the epoch angle of the resultant in Fig. 157 is ut2 =
toil — a, we can state the resultant as follows:
// we have given two S.H.M.'s in quadrature and take the amplitude
oj the one possessing the greater epoch angle as c and the amplitude
of the other S.H.M. as d, and construct the complex number c + di,
then this complex number c + di completely characterizes both of the
S.H.M.'s and their resultant. For, we can determine the modulus p
and the amplitude aoi c + di and then if wti is the epoch angle of the
moticm with amplitude c, the epoch angle of the resultant is ati — a.
If we consider the two harmonic
P motions
iEc p — 0,1 sin ui{t — ti)
and
p = 02 sin ci)(t — ti),
, ' and if Ji be greater than ti, the first
Fig. 158.— Complex number S.H.M. reaches its maximum value
representation of the facts ji^ ^^e second reaches its maxi-
shown by polar diagram, Fig. ^ mi. ^ ^ o ti at • ^i.
147 ^ B > B mum. The first S.H.M. is there-
fore said to lag the amount (<i — tij
behind the second S.H.M. That is, a S.H.M. represented by a circle
located anticlockwise from a second circle represents a S.H.M. that
lags behind the second.
220.* Illustration from Alternating Currents. The steady current
C flowing in a simple electric circuit is determined by the pressure or,
electromotive force, E and the resistance R according to the equation
known as Ohm's law,
^ r'
or
E = CR.
E is the pressure or voltage required to make the current C flow
against the resistance R. If the current, instead of being steady,
varies or fluctuates, then the pressure CR required to make the current
C flow over the true resistance is called the ohmic voltage, or ohmic
pressure. But a changing or fluctuating current in an inductive
circuit sets up a changing magnetic field around the circuit, from which
there results a counter electromotive force, or choking effect, due to the
changing of the current strength. This electromotive force is called
§220] COMPLEX NUMBERS 385
the reactive voltage or reactive pressure. The choking effect that it
has on the current is known as the inductive reactance. In case of a
periodically changing current it acts alternately with and against the
current. Opposite to the reactive voltage there is a component of the
impressed voltage that is consumed by the reactance. See Fig. 159.
The pressure which is at every instant applied to the circuit
from without is called the impressed electromotive force, or voltage.
Of the three pressures — namely, the impressed voltage, the ohmic
voltage (consumed by the resistance) and the reactive voltage con-
sumed by the inductive reactance — any one may always be regarded
as the resultajit of the other two. Hence, if in a polar diagram the
pressures be represented in magnitude and relative phase by the sides
of a parallelogram, the impressed
voltage may be regarded as the
diagonal of a parallelogram of
which the other two pressures are
sides. Since, however, the re-
actance or the counter inductive
pressure depends upon the rate of
change of the current, it lags, in
the case of a sinusoidal current,, Fiq. 159. — Complex number
90° behind the true, or ohmic, diagram of equation 5, §220.
voltage, which last is always in
phase with the current. The pressure consumed by the counter in-
ductive pressure therefore leads the current by 90°. Thus, in the
language of complex numbers
Ei=E„ + iE„ (1)
in which
Ei = impressed pressure
Eo = ohmic pressure, or pressure consumed by the resistance
Ec = counter inductive pressure, or the pressure consumed
by reactance.
It is found that the counter inductive pressure depends upon a con-
stant of the circuit L called the inductance and upon the angular veloc-
ity or frequency of the alternating impressed pressure, so that
E^ = 2ir/LC = wLC
Hence (1) may be written
Ei = RC + i2irfLC (2)
= flC + wLC (3)
386 ELEMENTARY MATHEMATICAL ANALYSIS [§220
The modulus of the complex number on the right of this equation is
Considering, then, merely the absolute value \Ec\ and \C\ of pressure
and current, we may write
Id = , '^'1 „ (4)
From the analogy of this to Ohm's law,
^ r'
the denominator -\/i22 _|. ^■iii ig thought of as limiting or restricting
the current and is called the impedance of the circuit.
Let there be a condenser in the circuit of an alternator, but let the
circuit be free from inductance. Then besides the pressure con-
sumed by the resistance, an additional pressure is required at any
instant to hold the charge on the condenser. If K be the capacity
of the condenser, it is found that that part of the pressure consumed
C C
in holding the charge on the condenser is ?r7v-' °^ ~xr' ^nd is in phase
position 90° behind the current C. The choking effect of this on the
current may be called the condensive reactance. When a condenser
is in the circuit in addition to inductance, the total pressure con-
sumed by the reactance has the form :
^"■^^ " 2^'
and the complex number that symbolizes the vector is
iP
Ei =RC + i2^fCL - ~. (5)
(see Fig. 159).
Further illustrations of the applications of complex numbers to
alternating currents is out of place in this book. The illustrations are
merely for the purpose of emphasizing the usefulness of these numbers
in applied science. An interesting application of the use of complex
numbers to the problem of the steam turbine will be found in Stein-
metz's "Engineering Mathematics," Page 33.
Exercises
1. Draw the polar diagram and complex number representation of
& if iJ = 5, C = 21,/ = 60, L = 0.009, K = 0.005.
2. Draw a similar diagram if Ei = 100, Eo = .90, / = 40, L = 0 . 008,
K = 0.003.
CHAPTER XIII
LOCI
221. Parametric Equations. The equation of a plane curve is
ordinarily given by an equation in two variables, as has been
amply illustrated by numerous examples in the preceding chapters.
It is obvious that a curve might also be given by two equations
containing three variables, for if the third variable be eliminated
from the two equations, a single equation in two variables results.
When it is desirable to describe a locus by means of two equations in
three variables the equations are known as parametric equations,
as has already been explained in §84. Two of the variables usu-
ally belong to one of the common coordinate systems and the
third is an extra variable called the parameter. In applied science
the variable time frequently occurs as a parameter.
The parametric equations of the circle have already been writ-
ten. They are
X = a cos d, (1)
y = a sin 6,
where the parameter d is the direction angle of the radius vector
to the point (a;, y). Likewise the parametric equations of the
elUpse have been written
X = a cos d, (2)
y r h sin d,
and those of the hyperbola have been written
a; = a sec d, (3)
y = h tan 0.
In harmonic motion, the ellipse was seen to be the resultant of
the two S.H.M. in quadrature
X = a cos (lit, (4)
2/ = 6 sin cat.
Here the parameter t is time.
387
388 ELEMENTARY MATHEMATICAL ANALYSIS [§222
222. Problems in Loci. It is frequently required to find the
equation of a locus when a description of the process of its genera-
tion is given in words, or when a mechanism by means of which the
curve is generated is fully described. There is only one way to
gain facility in obtaining the equations of curves thus described,
and that is by the solution of numerous problems. Sometimes
it is best to seek the parametric equations of the curve, but
sometimes the ordinary polar or Cartesian equation can be ob-
tained directly. The following problems are illustrative:
Fig. 160. — Generation of so-called "elliptic motion."
(1) A straight line of constant length a + b moves with its ends
always sliding on two fixed lines at right angles to each other.
Find the equation of the curve described by any point of the
moving line. (See §84, exercise 23.)
In Fig. 160, let AB be the line of fixed length, and let it so move
that A remains on the Z-axis and B remains on the F-axis. Let
any point of this line be P whose distance from A is 6 and whose
distance from B is a. If the angle X'AB be called 6, then PD,
the ordinate of P, is
y = b sin 9
and OD, the abscissa of P, is
X = a cos 6,
Therefore P describes an ellipse of semi-axes a and 6.
§222] LOCI 389
(2) A circle rolls without slipping within a circle of twice the
diameter. Show that any point attached to the moving circle'
describes an ellipse.
Draw the smaller rolling circle in any position within the larger
circle, and call the point of tangency T, as in Fig. 160. Since
the smaller circle is half the size of the larger circle, the smaller
circle always passes through 0, and the line Adjoining the points
of intersection of the small circle with the coordinate axes is, for
all positions, a diameter, since the angle AOB is a right angle.
If we can prove that the arc AT = the arc HT for all positions
of T, then we shall have shown that as the small circle rolls from an
initial position with point of contact at H, the end A of the diam-
eter AB slides on the line OX. Since B lies on OY and since AB
is of fixed length, this proves by problem (1) that any point of the
small circle lying on the particular diameter AB describes an
ellipse.
To prove that arc AT = arc HT, we have that the angle HOT
is measured in radians by „„ — • The angle AO'T is measured in
arc AT
radians by q,^ ■ Since Z AO'T = 2 Z HOT, we have
arc AT __ arc HT
O'A " OH '
But, OH = 20' A. Hence a.TC AT = arc HT.
We can now prove that any other point of the rolling circle de-
scribes an ellipse. Let any other point be Pi. Through Pi draw
the diameter JO'K. The above reasoning applies directly, re-
placing A by J and H by N.
It is easy to see that all points equidistant from the center of the
small circle, such as the points P, and Pi, describe ellipses of the
same semi-axes a and 6, but with their major axes variously in-
clined to OH.
(3) Determine the curve given by the parametric equations
X = a cos 2ut (1)
y = a sin ut. (2)
^"Circle" is here used in the sense of a "disc" or circular area and not in the
Beose of a " oiroumf erence."
390 ELEMENTARY MATHEMATICAL ANALYSIS [§222
To eliminate t, the first equation may be written
K = a (1 - 2sin2a)0. (3)
From the second equation, sin cot = ~. Substituting for sin ut in
(3),
- = «(l-^> (4)
or
2/^=-|^ + f- (5)
This curve is the parabola y^ = mx, the special location of ■which
the student should describe.
(4) Construct a graph such that the increase in y varies directly
as X.
If y varied directly as x, then y would equal kx, where A; is any
constant. In the given problem, however, the increase in y (and
not y itself) must vary in this manner. Let the initial value of y
be represented by z/o. Then the gain or increase of y is repre-
sented by 2/ — 2/0. Hence, by the problem,
y — yo = kx. (1)
Since t/o is a constant, (1) is the equation of the straight line of
slope k and F-intercept j/o, which ordinarily would be written in
the form
y = kx + 2/0.
(5) Express the diagonal of a cube as a function of its edge, and
graph the function.
If the edge of the cube be x, its diagonal is -v/a;" + x^ + x^ or
X \/3. If the diagonal be represented by y, we have y =\/3x,
which is a straight line.
(6) A rectangle whose length is twice its breadth is to be in-
scribed in a circle of radius a. Express the area of this rectangle
in terms of the radius of the circle.
Let the rectangle be drawn in a circle whose equation is
x^ + y^ = a'. At a corner of the rectangle we have x = 2y. The
area A of the rectangle is 4xy, or 8y^ since x = 2y. From the
§222] LOCI 391
equation of the circle we obtain 4y^ + y^ = a^ or y^ = a^/5.
Hence
A = (8/5)a2.
If A and a be graphed as Cartesian variables, the graph is a
parabola.
(7) A rectangle is inscribed in a circle. Express the area of the
rectangle as a function of a half of one side.
Here, as above,
A = 4x2/ = 4x Va^ — x^-
The student should graph this curve, for which purpose a may be
put equal to unity. First draw the semicircle y = \/'o^ — x'-
For X = 1/6, take one-fifth of the ordinate of this semicircle. For
X = 2/5, take two-fifths of the ordinate of the semicircle, and so on.
The curve through these points is y = x s/ a^ — x^, from which
y = 4x \/a^ — x'' can be had by proper change in the vertical unit
of measure.
Exercises
1. In polar coordinates, draw the curves:
p = 2 cos 8 p = 2 cos 9 + 1
P = 2 cos 9 — 1 p = 2 cos 9 -1- 3.
2. On polar coordinate paper select the point (1, 1). (This means
the point whose coordinates are one centimeter, and one radian.)
Starting at this point, a point moves so that the radius vector of the
moving point is always equal to the vectorial angle. Sketch the
curve. Write the polar equation of the curve.
3. A point moves so that one of its polar coordinates, the radius
vector, varies directly as the other polar coordinate, the vectorial
angle. Write the polar equation of such a curve. Does the curve
go through the point (!', 1)?
4. A polar curve is generated by a point Which starts at the point
(1, 2) and moves so that the increase in the radius vector always
equals the increase in the vectorial angle. Write the equation of the
curve.
6. A polar curve is generated by a point which starts at the point
(1, 2) and moves so that the increase in the radius vector varies directly
as the increase in the vectorial angle. Write the equation of the curve.
6. A ball is thrown from a tower with a horizontal velocity of 10
392 ELEMENTARY MATHEMATICAL ANALYSIS [§222
feet per second. It falls at the same time through a variable distance
given by s = 16. 1<^, where t is the elapsed time in seconds and a is
in feet. Find the equation of the curve traced by the ball.
7. The point P divides the line AB, of fixed length, externally in
the ratio a : 6, that is, so placed that PA/PB = a/b. If the line AB
move with its end points always remaining on two fixed lines OX and
OK at right angles to each other, then P describes an ellipse of semi-
axes a and b.
8. If in the last problem the lines OX and OY are not at right
angles to each other, the point P still describes an ellipse.
9. A point moves so as to keep the ratio of its distances from two
fixed lines AC and BD constant. Prove that the locus consists of
four straight hnes.
10. A sinusoidal wave of amplitude 6 cm. has a node at + 5 cm.
and an adjacent crest at + 8 cm. Write the equation of the curve.
11. The velocity of a simple wave is 10 meters per second. The
period is two seconds. Find the wave length and the frequency.
12. A polar curve passes through the point (1, 1) and the radius
vector varies inversely as the vectorial angle. Plot the curve and
write its equation. Consider especially the points where the vectorial
angle becomes infinite and where it is zero. Sketch the same func-
tion in rectangular coordinates.
13. Rectangles are inscribed in a circle of radius r. Express by
means of an equation and plot: (o) the area, and (6) the perimeter of
the rectangles as a function of the breadth.
14. Right triangles are constructed on a line of given length h as
hypotenuse. Express and plot: (a) the area, and (6) the perimeter as
a function of the length of one leg.
16. A conical tent is to be constructed of given volume, V. Express
and graph the amount of canvas required as a function of the radius
of the base.
16. A closed cylindrical tin can is to be constructed of given volume,
V. Plot the amount of tin required as a function of the radius of the
can.
17. A rectangular water-tank lined with lead is to be constructed
to hold 108 cubic feet. It has a square base and open top. Plot
the amount of lead required as a function of the side of the base.
18. An open cylindrical water-tank is to be made of given volume,
V. The cost of the sides per square foot is two-thirds the cost of the
bottom per square foot. Plot the cost as a function of the diameter.
19. An open box is to be made from a sheet of pasteboard 12 inches
square, by cutting equal squares from the four comers and bending up
§223]
LOCI
393
the sides. Plot the volume as a function of the side of one of the
squares out out.
20. The illumination of a plane surface by a luminous point varies
directly as the cosine of the angle of incidence, and inversely as the
square of the distance from the surface. Plot the illumination of a
point on the floor 10 feet from the wall, as a function of the height
of a gas burner on the wall.
21. Using the vertical distances between corresponding points on
the curves y = sin t and y = — sin t as ordinates and the vertical
distances between corresponding points oi y = 2t and j/ = t^ as abscis-
sas, find the equation of the resulting curve.
223. Loci Defined by Focal Radii. A number of important
curves are defined by imposing conditions upon the distances of
any point of the locus from two fixed points, called foci.
Pig. 161. — The lepmiscate.
(1) A point moves so that the product of its distances from two
fixed points is constant. Find the equation of the path of the par-
ticle. Let the two fixed points Fi and Fi, Kg. 161, be taken on the
X-axis the distance o each side of the origin. Call the distances of P
from the fixed points ri and rj. Then the variables ri and rj in terms
of ::; and y are
ri'^ = y' + (x - a)2
Hence
ri' =y' + (x + a)2.
nW = [y^ + (a; - o)"] [y^ + {x + a)'].
(1)
(2)
Calling the constant value of tiTi = c', we have, as the Cartesian
equation of the locus,
y' + (x- ay] [y^ + (x + a)»] = c\
(3)
394 ELEMENTARY MATHEMATICAL ANALYSIS [§223
Fig.
162. — The lemniscate and the
Cassinian ovals.
which may be written
(2/2 + a;2 + a2)2 - 4a^'x^ = c* (4)
(x2 + y'y + 2aV + 20^2/2 + a* - iaV = c* (5)
V (a;2 + 2/2)2 ^ 2a%x' - y') + c* - a'. (6)
If c = a the curve is called
the lemniscate, and the Car-
tesian equation reduces to
(x2+ 2/2)2 = 2a2(a;2 - 2/=^)- (7)
For other values of c the
curves are known as the
Cassinian ovals. The stu-
dent will show that when
c < u, the curve consists of
two separate ovals surround-
ing the foci, and for c > a
there is but a single oval.
The curves are shown in Fig. 162. These curves give the form of
the equipotential surfaces in a field around two positively or two
negatively charged parallel wires.
(2) Construct the curve such that the ratio of the distances of any
of the curve from two fixed points is constant. Let the two fixed
points be A and B, Fig.
163; let the constant ratio
of the distances of any point
of the curve from the two
fixed points be n/J'2 = mm.
To find one point of the
locus, draw circles from A
and B as centers whose radii
are in the ratio m/n. Let
these circles intersect at the
point P. At P bisect the
angle between PA and PB
internally and externally by
the lines PM and PN respectively. The line AB ia then divided
at M internally in the ratio MA/ MB = m/n and externally at N
in the ratio NA/NB = m/n, because the bisectors of any angle of
a triangle divide the base into segments proportional to the adja-
cent sides. Since the external and internal bisectors of any angle
Fig. 163. — Construction of 'the curve
ri/Ti = m/n, or the circle MPN.
§224] LOCI 395
must be at right angles to each other, PM is perpendicular to PiV
for any position of P. Hence the locus of P is a circle, since it is the
vertex of a right triangle described on the fixed hypotenuse MN.
If a large number of circles be drawn for different values of m/n, and
if similar circles be described about B, then these circles are known
as the dipolar circles. See Fig. 164. In physics it is found that these
circles are the equipotential hnes about two parallel wires perpendic-
ular to the plane of the paper at A and B and carrying electricity of
opposite sign.
Fig. 164. — The dipolar circles, or a family of circles made by
drawing ri/ra = e for various values of e.
Exercises
1. Draw the locus satisfying the condition that the ratio of the
distances of any point from two fixed points ten units apart is 2/3.
2. Draw the two circles which divide a line of length 14 internally
and externally in the ratio 3/4.
224. The Cycloid. The cycloid is the curve traced by a point
on the circumference of a circle, called the generating circle,
which rolls without slipping on a fixed line called the base. To
find the equation of the cycloid, let OA,'Yig. 165, be the base, P
the tracing point of the generating circle in any one position, and
Q the angle between the radii SP and SH. Since P was at 0 when
the circle began to roll,
OH = ad,
396 ELEMENTARY MATHEMATICAL ANALYSIS [§225
if a be the radius of the generating circle. Since »= OD and
y = DP, we have
x= OH- SP sin e= a(e - sin 6)
y= HS- SP cos e= a(i- cos (9).
(1)
(2)
These ar^ the parametric equations of the curve. For most
purposes these are more useful than the Cartesian equation.
o D H c A -
Fig. 165. — Definition of the cycloid.
It is readily seen from the definition of the curve, that the locus
consists of an unlimited number of loops above the X-axis, with
points of contact with the X-axis at intervals of 2xa (the circum-
ference of the generating circle) and with maximum points at
X - Ta, 3xo, etc.
"=*'2Pl6 12 3 4 6 C A
Fig. 166. — Construction of the cycloid.
225.* Graphical Construction of the Cycloid. To construct the
cycloid, Fig. 166, draw a circle of radius 1.15 inches and divide the
circumference into thirty-six equal parts. Draw horizontal lines
through each point of division exactly as in the construction of
the sinusoid, Fig. 59- Lay off uniform intervals of 1/5 inch each
on the X-axis, marked 1, 2, 3, . . . Then from the point of
division of the circle pi lay off the distance 01 to the right.
§226]
LOCI
397
From pi lay off 02 to the right, from pa lay off 03 to the right,
etc. The points thus determined lie on the cycloid. The number
of divisions of the circumference is of course immaterial except
that an even number of division is .convenient. Further the
divisions laid off on the base OA must be the same length as
the arcs laid off on the circle.
Note that by the process of construction above, the vertical
distances from OX to points on the curve are proportional to
(1 — cos 6) and that the horizontal distances from OY to points
on the curve are proportional to (d— sin 6).
The analogy of the cycloid to the sine curve is brought out by
Fig. 167. A set of horizontal lines are drawn as before and also a
sequence of semicircles spaced at horizontal intervals equal to
Fig. 167.— Analogy of the cycloid to the sinusoid.
the intervals of arc on the circle. The plane is thus divided
into a large number of small quadrilaterals having two sides
straight and two sides curved. Starting at 0 and sketching the
diagonals of successive cornering quadrilaterals the cycloid is
traced. If, instead of the sequence of circles, uniformly spaced
vertical straight lines had been used, the sinusoid would have been
drawn; The sinusoid on that account is frequently called the
"companion to the cycloid."
226. Epicycloids and Hj^ocycloids. The curve traced by a
point attached to the circumference of a circle which rolls without
slipping on the circimiference of a fixed circle is called an epi-
cycloid or a hypocycloid according as the rolling circle touches
the outside or inside of the fixed circle. If the tracing point
is not on the circumference of the rolling circle but on a radius
or radius produced, the curve it describes is called a trochoid if
398 ELEMENTARY MATHEMATICAL ANALYSIS [§226
the circle rolls upon a straight line, or an epitrochoid or a hypo-
trochoid if the circle rolls upon another circle.
Exercises
1. Construct a cycloid by dividing a generating circle of radius
1.15 inches into twenty-four equal arcs and dividing the base into
intervals 3/10 inch each.
2. Compare the cycloid of length 2ir and height 1 with a semi-
ellipse of length 2ir and height 1.
3. Write the parametric equations of a cycloid for origin C, Fig. 165.
4. Write the parametric equations of a cycloid for origin B, Fig. 165.
6. Show that the top of a rolling wheel travels through space twice
as fast as the hub of the wheel.
6. By experiment or otherwise show that the tangent to the cycloid
at any point always passes through the highest point of the generating
circle in the instantaneous position of the circle pertaining to that
point.
CHAPTER XIV
THE CONIC SECTIONS
227. The Focal Radii of the EUipse. Draw any ellipse with
major and minor circles of radii a and 6 respectively, as in Fig.
168. Draw tangents, II' and KK', to the minor circle at the
extremities of the minor axes and comJ)lete the rectangle II'KK'.
Properties of the elUpse.
The points Fi and Fi, in which IK and I'K' cut the major axis, are
called the foci of the ellipse. Prom any point on the ellipse draw
the focal radii PFj = rj. and PF2 = r^, as shown in the figure.
Represent the distance OFi = OF 2 by c. Then it follows from
the triangle OIFi that
a^ = b« + c^. (1)
This is one of the fundamental relations between the constants of
the ellipse. >
399
400 ELEMENTARY MATHEMATICAL ANALYSIS [§227
From the triangles PFiD and PFiD there follows:
n^ = (c - xY + v' (2)
r/ = (c + xy + if. (3)
But the equation of the ellipse is
b ,
2/ = - Vo' - x\
or
y^ = ^,(a^ - a;^). (4)
Substituting this value of y^ in (2)
ri' = c'' - 2cx + x^ + -.(a'' - x^) (5)
= c^ -2cx + x^ + ¥ -~i a;^
or by (1)
r," = a^ - 2cx + a;^ [l - -^J . (6)
Substituting
1 _ ^ - «' - fe' _ ^,
we obtain
/i2/»2
ri2 = o2 - 2ca; + ^ (7)
= [•-"]"" <»)
Therefore
Likewise, from (3), by exactly the same substitutions, there
follows
r2 = a + "''■ (10)
a
From (9) and (10), by addition,
ri + Tz = 2a. (11)
Hence in any ellipse the sum of the focal radii is constant and equal
to the major axis,
§228] THE CONIC SECTIONS 401
The converse of this theorem, namely, if the sum of the focal
radii of any locus is constant, the curve is an ellipse, can readily
be proved. It is merely necessary to substitute the values of ri
and Ti from (2) and (3) in equation (11), and simplify the resulting
equation in x and y; or first square (11) and then substitute ri and
r-2 from (2) and (3). There results an equation of the second
degree lacking the term xy and having the terms containing x'' and
y^ both present and with coefficients of like signs. By §86, such an
equation represents an ellipse.
Hence the ellipse might have been defined as the locus of a
point, the sum of whose distances from two fixed points is constant.
An ellipse can be drawn by attaching a string of length 2a by
pins at the points F\ and F2 and tracing the curve by a pencil so
guided that the string is always kept taut. Or better, take a
string of length 2a + 2c and form a loop enclosing the two pins;
the entire curve can then be drawn with one sweep of the pencil.
The focal radii may also be evaluated in terms of the para-
metric or eccentric angle 0. The student may regard the follow-
ing demonstration of the truth of equation (11) as simpler than
that given above.
Since a; = a cos 6, and y = h sm d
equation (2) gives
ri" =¥ sin^ e -h (c - a cos df (12)
= h^aia^e + c^ - 2accos0 + o^ cos^ d. (13)
To put the right side in the form of a perfect square, write
W = a^ - c^. Then
ri' = a^ sin'' 0 — c^ sin'' B + c^ — 2ac cos d + a' cos^ 6
= o^ - 2oc cos e +c'^ cos^ e. (14)
Whence
Likewise
Whence
ri = a — c cos 9. (15)
rj = a -|- c cos 6. (16)
ri -H r2 = 2a.
228. The Eccentricity. The ratio c/a measures, in terms of o as
unit, the distance Pf either fopus from the center of the ellipse.
26
402 ELEMENTARY MATHEMATICAL ANALYSIS [§229
This ratio is called the eccentricity of the ellipse. In the triangle
IFiO, the ratio c/a is the cosine of the angle FiOI, represented in
what follows by /3. Calling the eccentricity e, we have
e = c/a = cos |8. (1)
The ellipse is made from the major circle by contracting its ordi-
nates in the ratio m = b/a, or by orthographic projection of the
circle through the angle of projection
a = cos~i b/a.
Hence, as companion to (1) we may write
m = b/a = cos a = sin |8. (2)
229. The Ratio Definition of the Ellipse. In Fig. 168, let the
tangents to the major circle at I and I' be drawn. Draw a
perpendicular to the major axis produced at the points cut by
these tangents. These two lines ai'e called the directrices of the
ellipse.
We shall prove that the ratio PFi/PH (or PFt/PH') is constant
for all positions of P. From §227, equation (9) or (15),
ri = a — c cos d, (1)
/?• (2)
(3)
From the figure,
ON = a sec ION = a
But
cos |8 = c/a.
Hence,
ON = a^/c.
But
PH = 0N - X.
Therefore
PH = a'^/c - a cos 6.
Hence, from (1) and (4),
r\
= PI
^ /PH - "'-'' cos 9
PH
a'^/c — acos e
c a — c cos 6
(4)
a a— c cos d
or
PFi/PH ^ c/a = e = cos |S. . , , . (6)
§229] THE CONIC SECTIONS 403
A similar proof holds for the other focus and directrix. Thus,
for any point on the ellipse the distance to a focus bears a fixed
ratio to the distance to the corresponding directrix. From (5),
the ratio is seen to be less than unity.
Assuming the converse of the above, the ellipse might have been
defined as follows : The ellipse is the locus of a point whose distance
from a fixed point (called the focus) is in a constant ratio, less than
unity, to its distance from a fixed line (called the directrix).
If, in any ellipse, c = 0, it follows that b must equal a and the el-
lipse reduces to a circle. If c is nearly equal to a, then from the
equation
a2 = 62 + c^,
it foljows that the semi-minor axis & must be very small. That is,
for an eccentricity nearly unity the ellipse is very slender.
If the sun be regarded as fixed in space, then the orbits of the
planets are ellipses, with the sun at one fobus. (This is "Kepler's
First Law.") The eccentricity of the earth's orbit is 0.017. The
orbit of Mercury has an eccentricity of about 0.2, which is greater
than that of any other planet.
Exercises
Find the eccentricities and the distance from center to foci of the
following elUpses:
1. a;V9 -I- 2/74 = 1. i. 2y = Vl - x".
2. 2/ = (2/3.)-\/36 - xK 5. Qx" + IQy' = 14.
3. 25a;2 + iy' = 100. 6..2a;2 + 3^^ = 1.
Find the equation of the ellipse from the following data:
7. e = 1/2, a = 4. Draw this ellipse.
8. c = 4, a = .5.
9. ri = 6 - 2a;/3, ri = 6 + 2a;/3.
10. ri = 5 — 4 cos 0, ri = 5 -j- 4: cos 6.
Solve the following exercises:
11. Find the eccentricity of the ellipse made by the orthographic
projection of the circle x' + y^ = a' through the angle 60°.
, 12. The angle of projection of a circle x' + y' = a' by which an
404 ELEMENTARY MATHEMATICAL ANALYSIS [§230
ellipse is formed is a. Show that the eccentricity of the ellipse is
sia a.
13. A circular cylinder of radius 5 is cut by a plane making an
angle 30° with the axis. Find the eccentricity of the elliptic section.
14. If the greatest distance of the earth from the sun is 92,500,000
miles, find its least distance. (Eccentricity of earth's orbit = 0.017.)
16. In the ellipse a;*/25 + y'/16 = 1, find the distance between
the two directrices.
16. Write the equation of the ellipse whose foci are (2, 0), ( — 2, 0),
and whose directrices are x = 5 and a; = — 5.
17. Prove equation 11 §227 by transposing one radical in:
V(.x+c)^+y^ + V{x - c)2 + y' = 2o
squaring, and reducing to an identity.
18. Obtain the equation of the ellipse from the definition at the top
of page 403.
230. The Latus Rectum. The double ordinate through the
focus is called the latus rectum of the ellipse. The value of the
semi-latus rectum is readily formed from the equation
y = (b/a) Va" - a;'
by substituting c for x. If I represents the corresponding value
oiy,
I = (b/a) Va^ - c2 = 6Va (1)
(2)
since a^ — c'^
= 6^.
Hence the entire
lnius
by
21 = ?^^
a
Equation (1)
may
be also be written
l = bVi-
c-'/a-
= b Vi-
e\
(3)
In Fig. 168 the distances AF, AN, ON, OB, OF, FN may
readily be expressed in terms of a and e as follows in equations (4)
to (10). The addition of the formulas (11), (12), (13) brings into
a single table all the important formulas of the ellipse.
AFi = a - c = a(i - e) (4)
§230]
THE CONIC SECTIONS
405
e e
(5)
ON = a sec ^ = 1
(6)
e = cos |8
(7)
OB = b = a sin /3 = a Vi - e^
(8)
OFi = c = ae
(9)
FiN = OiV - c = a(i - e=)/e
(10)
1 = bVa = a(i - e'^)
(11)
Ti = a — ex = a — X cos |3
(12)
ra = a + ex = a + X cos j3
(13)
Exercises
1. Find the value in miles of OFi for the case of the earth's orbit.
2. Find the equation of an ellipse whose minor axis is 10 units and
in which the distance between the foci is 10.
3. In the ellipse y = (2/3)\/36 - x' find the length of the latus
rectum and the value of e.
4. The eccentricity of an ellipse is 3/5 and the latus rectum is 9
units. Find the equation of the ellipse.
6. In (a) X' + 4v2 = 4 and (6) 2x' -|- 32/^ = 6 find the latus rfectum,
the eccentricity, and the distances ON and AF.
6. Determine the eccentricities of the ellipses,
(o) j/2 = 4s - (l/2)x2 (b) j/2 = 4x - 2x\ .
7. Find the value of /3 for the earth's orbit. (Use the S functions
of the logarithmic table.)
8. Find the equation of an ellipse whose latus rectum is 2 units and
minor axis is 4.
9. The distance from the focus to the directrix is 16 units. The
distance from the vertex to the nearest focus is 6 units. Find the
equation of the ellipse.
10. The axes of an ellipse are known. Show how to locate the foci.
11. In an ellipse a = 25 feet, e = 0.96. What are the values of
c and 6?
12. For a certain comet (Tempel's) the semi-major axis of the'
elliptic orbit is 3.5, and c = 1.4 on a certain scale. For another
406 ELEMENTARY MATHEMATICAL ANALYSIS [§231
comet (Enke's) o = 2.2, e = 0.85. Sketch the curves, taking 3 cm.
or 1 inch as unit of measure.
13. If i = 7.2 and e = 0.6, find c, a, 6.
14. An ellipse, with center at the origin and major axis coinciding
with the X-axis, passes through the points (8, 3) and (6, 4). Find the
axes of the ellipse.
231. Focal Radii of the Hjrperbola. Construct a hyperbola
from auxiliary circles of radii o and b, then the transverse axis of
the hyperbola is 2a and the conjugate axis is 26. Unlike the case
of the ellipse, b may be either greater or less than a. As previously
explained, the asymptotes are the extensions of the diagonals of the
rectangles BTAO, BT'A'O. From the points /, /', in which the
\^
H
nX
t'
Hy^
W^
f
i
r
\m
y/
Fo .
A'
\
V
>y
A
I
^Fi
D
y^
(2)
B' ^y
<
Fig. 169.— Properties of the hyperbola.
asymptotes cut the a-circle, draw tangents to the o-circle. The
points Fi, F2 in which the tangents cut the axis of the hyperbola
are called the foci. See Fig. 169.
The distance OFi or OFa is represented by the letter c. Then,
since the triangles FJO and OAT are equal, FJ must equal 6, so
that we Jbave the fundamental relation between the constants of
"the hyperbola
a^ + b^ = c"
(1)
§232] THE CONIC SECTIONS 407
From any point on either branch of the hyperbola draw the focal
radii PFi and PF2, represented by ri and rj respectively. Then
from the figure
ri^ = (x - c)'+y\ (2)
But from the equation of the hyperbola
y^ = {V/a?){z^ - a?), (3)
hence
n^ = (x - cY + Wix" - a?) la"- (4)
= {aH^ - 2a^cx + aH^ + Vx^ - a^V") /a" (5)
= {cH'' - 2a?ex + a") /a?- (6)
= {ex - o?)ya\ (7)
Hence r\ = {c/a)x — a. (8)
In like manner it may be shown that
r2 = {c/a)x + a. (9)
Hence from (8) and (9) it follows
r2 — Ti = 2a. (10)
Thus^ in any hyperbola, the difference between the distances of any
point on it from the foci is constant and equal to the transverse axis.
The above relation may be derived in terms of the parametric
angle 9. Since in any hyperbola x = a sec 8 and y = b tan 6,
ri" = V tan'' (9 + (o sec 0 - cf
= W tan^ 6 + a' sec" 9 — 2ac sec 0 + c\
To put the right-hand side in the form of a perfect square, write
¥ = c^ - a\ Then
ri^ = c' sec'* 9 - 2ac sec 9 + a\
Therefore
ri = 0 sec 9 — a.
and
I2 = c sec 9 + a.
(11)
(12)
232. The Ratio Property of the Hyperbola. Through the
points of intersection of the a-circle with the asymptotes, draw
408 ELEMENTAKY MATHEMATICAL ANALYSIS (§233
IK, and I'K'. These lines are called the diiectrices of the
hyperbola. It wiU now be proved that the ratio of the distance of
any point of the hyperbola frota a focus to its distance from the
corresponding directrix is constant. Adopt the notation
c/a = sec ;8 = e. (1)
Then, from the figure
PFi/PH = ri/{x - ON) = ri/(a sec 0 - o cos ;8) (2)
Substituting ri from (11) above:
PF^/PH = "^'l^-'* ^ (3)
' a sec 0 — a cos j8 ^ '
= c sec 8 — a c
(4)
a (^sec e - -j
which proves the theorem. The constant ratio e is called the
eccentricity of the hyperbola, and, as shown by (1), is always
greater than unity.
Assuming the converse of the above, it is obvious that the hyper-
bola might have been defined as follows: The hyperbola is the
locus of a point whose distance from a fixed point (called the focus)
is in a constant ratio, greater than unity, to its distance from a fixed
line {called the directrix). Proof will be given in §234.
233. The Latus Rectum. The double ordinate through the
focus is called the latus rectum of the hyperbola. The value of
the semi-latus rectum is readily found from the equation
y = (b/a) Vx^ - a^
by snbstituting c f or a;. HI represent the corresponding value of y,
I = (b/a) Vc' - a^ = hya. (1)
Hence the entire latus rectum is represented by
2l = 2bVa. (2)
Equation (1) may also be written
i = 6 Ve' - 1. (3)
§233]
THE CONIC SECTIONS
409
In Fig. 169 the distances AFi, AN, ON, OB, OFi, FiN may
readily be expressed in terms of o and e, as follows in equations
(4) to (8). Collecting in a single table the other important for-
mulas for the hyperbola, we have :
AFi = c — a = a(e — i)
(4)
AN = AF^/e = a(e - i)/e
(5)
ON = a cos /3 = a/e
(6)
e = sec /3
(7)
OB = b=atan/3 = a Ve^ - i
(8)
OFi = c = ae
FiN
= c - OiV = ae - a/e = aCe^ - i)/e
(9)
1 = bVa = b Ve^ - 1 = a(e2 - i)
(10)
ri = ex — a = X sec iS — a
(11)
Tz = ex + a = X sec |3 + a
(12)
The important properties of the hyperbola are quite similar
to those of the ellipse. It is a good plan to compare them in
parallel columns.
Ellipse
Hyperbola
1. Definition of Foci and Focal
Radii
2. a2 = b2 + c2
3. ri + ra = 2a
4. Eccentricity, e = - = cos /3
5. Definition of Directrices
PFi
6. The Ratio Property, -pg = e
262
7. The Latus Rectum = —
1. Definition of Foci and Focal
Radii
2. a^ + b-' = c2
3. ra — ri = 2a
4. Eccentricity, e = - = sec /3
5. Definition of Directrices
PFi
6. The Ratio Property, p^ = e
2b^
7. The Latus Rectum = —
410 ELEMENTARY MATHEMATICAL ANALYSIS [§234
Exercises
1. Find the eccentricity and axes of sV^ — y'/^^ = 1-
2. Find the eccentricity and latus rectum of the hyperbola con-
jugate to the hyperbola of the preceding exercise.
3. A hyperbola has a transverse axis equal to 14 units and its
asymptotes make an angle of 30° with the Z-axis. Find the equation
of the hyperbola.
4. Find the latus rectum and locate the foci and asymptotes of
4i2 - 362/2 = 144.
6. Locate the directrices of the hyperbola of the preceding exercise.
6. In Fig. 169 show that rz = GK' and ri = GI and hence that
rj — ri = IK' or 2a.
7. Find the equation of the hyperbola having latus rectum 4/3
and a = 26.
8. The eccentricity of a hyperbola is 3/2 and its directrices are the
lines X = 2 and x = — 2. Write the equation and draw the curve
with its asymptotes, a-circle, 6-circle, and foci.
9. Find the eccentricity and axes of 3x^ — 5y' = — 45.
10. Find the eccentricity of the rectangular hyperbola.
11. Describe the shape of a hyperbola whose eccentricity is nearly
unity. Describe the form of a hyperbola if the eccentricity is very
Large.
12. Describe the hyperbola if b/a = 2, but a very small.
13. Write the equation of the hyperbola if (1) c = 6, o = 3; (2)
c = .25, o = 24; (3) c = 17, 6 = 8. /
14. Describe the locus (,x + 1)V7 - iy - 3)V5 = 1.
15. Find the equation of the hyperbola whose center is at the origin
and whose transverse axis coincides with the X-axis and which passes
through the points (4.5, — 1), (6, 8).
234. The Polar Equation of the Ellipse and Hyperbola. In
mechanics and astronomy the polar equations of the ellipse and
hyperbola are often required with the pole or origin at the right-
hand focus in the case of the ellipse and at the left-hand focus in
the case of the hyperbola. In these positions the radius vector
of any point on the curve will increase with the vectorial angle
when B < 180°. To obtain the polar equation of the ellipse and
hyperbola, make use of the ratio property of the curves, namely:
That the locus of a point whose distances from a fixed point (called
the focus) is in a constant ratio e to its distances from a fixed
§234]
THE CONIC SECTIONS
411
line (called the directrix), is an ellipse if e < 1 or a hyperbola
if e > 1. In Fig. 170 let F be the fixed point, or focus, IK the
fixed line, or directrix, P the moving point, and FL = I the semi-
latus rectum. Then the problem is to find the polar equation
under the condition
pg-e (1)
If e is understood to be unrestricted in value, the work and
the result will apply equally well either to the ellipse or to the
hyperbola.
Fig. 170. — Polar equation of a conic.
When the point P occupies the position L, Fig. 170, we have
PF = I and PH = FN, whence from (1)
(2)
FN = ^-.
e
Take the origin of polar coordinates at F, and also take FP = p
and the angle AFP = 6. Then
PH = FN - FD (3)
FD = p cos e. (4)
412 ELEMENTARY MATHEMATICAL ANALYSIS [§234
Hence from (2), (3), and (4)
PH = - - p cos e. (5)
e
Substituting these values of FP and PH in (1), clearing of frac-
tions and solving for p, we obtain
" = I + e cos e ^^^
which is the equation required.
When e < 1, (6) is the equation of an ellipse with pole at the
right-hand focus. When e > 1, (6) is the equation of a hyperbola
with the pole at the left-hand focus. In both cases the origin has
been so selected that p increases as d increases.
Note: Calling FN (Fig. 170) = n, equation (1) above may be
written in rectangular coordinates
^^' + y' =e, (7)
n — X
x' + y^ = e\n - x)' ^°'
which may be reduced to the form
/ we' \ 2 y' _ e'n'
r "•■ 1 - eV "^ 1 - e' (1 -e')^' ^^'
By §§86 and 90 this represents an ellipse if e < 1 or a hyperbola
if e > 1. Thus starting with the ratio definition (7) we have proved
that the curve is an ellipse or a hyperbola; that is, we have proved
the statements in italics at end of §§229 and 232.
Exercises
n
1. Graph on polar paper, form M3, the curve p = j—r-^ i
for e = 2, e = 1/2, and e = 1.
It will be sufficient in graphing to use 9 = 0°, 30°, 60°, 90°, 120°,
160°, 180°, 210°, . ., 360°.
2. Write the polar equation of an ellipse whose semi-latus rectum
is 6 feet and whose eccentricity is 1/3.
3. Write the polar equation of an ellipse whose semi-axes are 5 and 3.
4. Discuss equation (6) for the case e = 0.
§235]
THE CONIC SECTIONS
413
6. Write the polar equation of a hyperbola if the eccentricity be
-\/2 and the distance from focus to vertex be 4.
6. Write the polar equation of the asymptotes of
4 + 5 cos
e
9
-and
p =
9
e'
4+5 cos
I
4
— 5 cos
in
which
a
is
1 +e cos
(9-
«)'
7. Compare the curves p =
8. Discuss the equation p =
constant.
235. Ratio Definition of the Parabola. Among the curves of the
parabolic type previously discussed, the one whose equation is of
Fig. 171. — Properties of the parabola y^ = ipx.
the second degree is of paramount importance. On that account
when the term parabola is used without qualification, it is under-
stood that the curve is the parabola of the second degree, whose
equation may be written, y^ = ax or x^ = ay.
We shall prove : The locus of a 'point whose distance from a fixed
point is always equal to its distance from a fixed line, is a parabola.
In Fig. 171, let F be the fixed point and HK the fixed line. Take
the origin at A half way between F and HK. Let P be any point
satisfying the condition PF == PH. Call OD = x, PD = y, and
represent the given distance FK by 2p. Then, from the right
triangle PFD,
414 ELEMENTARY MATHEMATICAX ANALYSIS [§230
ppi = 2/2 + FD^ (1)
= 2/2 + {x - OFY
= 1/2 + (a; - p)2.
Since PF by definition equals PH or a; + p, we have
{x + vY = 2/2 + {x - v)\ (2)
whence
y^ = 4px, (3)
which is the equation of the parabola in terms of the focal distance,
OF or p.
The double ordinate through F is called the latus rectum.
The semi-latus rectum can be obtained from (3) by placing
X = p, whence
1 = 2p, (4)
where / is the semi-latus rectum. Hence the entire latus rectum is
4p, or the coefficient of x in equation (3).
In Fig. 171, the quadrilateral FLIK is a square since FL and
FK are each equal to 2p.
236. Polar Equation of the Parabola. In accordance with the
ratio definition of the parabola, its polar equation is found at
once from equation (6), §234, by putting e = 1. Hence the polar
equation of the parabola is
'' = 7Tiosl" (^)
For this equation we may make the following table of values:
e
P
0°
1/2
90°
I
180°
00
270°
I
This shows that the parabola has the position shown in Fig. 171.
This is the form in which the polar equation of the parabola is
used in mechanics and astronomy.
237. The Conies. It is now obvious that a single definition
can be given that will include the ellipse, hyperbola and parabola.
§237] THE CONIC SECTIONS 415
These curves taken together are called the conies. The definition
fnay be worded: A conic is the locus of a point whose distances
from a fixed point (called the focus) and a fixed line (called the
directrix) are in a constant ratio. The unity between the three
curves was shown by their equation in polar coordinates. Moving
the ellipse so that its left vertex passes through the origin, as in
§85, and writing the hyperbola with the origin at the right ver-
tex (so that both curves pass through the origin in a comparable
manner), we may compare each with the parabola as follows:
TheeUipse: y^ = 2lx - (b^/a^)x'' (1)
The parabola: j/' = 2lx (2)
The hyperbola: y'' = 2lx + (b''/a')x'' (3)
In these equations I stands for the semi-latus rectum of each
of the curves. These equations may also be written
1/2 = 2lx - (l/a)x' (4)
2/^ = 2lx I (5)
2/2 = 2lx + (l/a)x'' (6)
whence it is seen that if Z be kept constant while a be increased
without limit, the ellipse and hyperbola each approach the parab-
ola as near as we please. Only for large values of x, if a be large,
is there a material difference in the shapes of the curves.
Exercises
1. Write the equation of the circle in the form (1) above.
2. Write the equation of the equilateral hyperbola in the form (3) ■
above.
3. Describe the curve
^ I
'' 1 -I- cos (e - a)'
where a is a constant.
4. In Fig. 172 translate the curve xy = Ihy suitable change in the
equation to the position shown by the dotted curve, if the translation
of each point is unity.
416 ELEMENTARY MATHEMATICAL ANALYSIS [§237
6. In Pig. 173 translate the curve j/* = 4px by suitable change in
the equation to the position shown by the dotted curve, if the distance
each point is moved be 3p.
Fig. 172. — A hyperbola translated at an angle of 45° to OX.
6. A bridge truss has the form of a circular segment, as shown in
Fig. 174. If the total span be 80 yards and the altitude BS be 20
Y
0^
j^
0
^-^^
X
— —i.
A l» Ci 10 Ca
Fig. 173. — A parabola translated Fig. 174. — Bridge truss in form
at an angle of 60° to OX. of circular segment.
yards, fmd the ordinates CiDi, C2D2, etc., erected at uniform intervals
of 10 yards along the chord AAi.
7. A bridge truss has the form of a parabolic segment, as shown in
Fig. 175. The span AAi is 24 yards and the altitude OB is 10
§238]
THE CONIC SECTIONS
417
erected at
yards. Find the length of the ordinates CD, CiDi, .
untform intervals of 3 yards along the Une AAi-
238.* The Conies are Conic Sections. The curves nowtnown as
the conies were originally studied by the Greek geometers as the sec-
X
3 6
9 12
•^'^^
OoS
\
\
Ai B
Y
Cs Ci C A
Fig. 175. — Bridge truss in the form of a paraboUc segment.
tions of a circular cone cut by a plane. It can be shown that the three
classes of curves, parabola, eUipse, and hyperbola, can be made
respectively by cutting any circular cone : (1) by a plane parallel to an
Fig. 176. — Section of a circular cone.
element; (2) by a plane cutting opposite elements of the same nappe of
the cone; (3) by a plane cutting both nappes of the cone. The two
nappes of a conical surface, it will be remembered, are the two portions
of the surface separated by the apex.
27
418 ELEMENTARY MATHEMATICAL ANALYSLS [§239
In Fig. 176, let the plane NDN'D', caUed the cutting plane, cut the
lower nappe of a right circular cone in the curve VPV. It can be
proved by geometry that this curve is an ellipse. The foci F and F
are the points at which the two inscribed spheres SFS' and RF'R' are
tangent to the plane ND'. The directrices are the two lines ND
and N'D' in which the plane ND' cuts the two planes SHS' and
RKR' produced.
239. Tangent to the Parabola. Let us investigate the condition
that the line y = mx + b shaU be tangent to the parabola y^ =
ipx. First find the points of intersection of these loci by solving
the two equations
y = mx + 6 (1)
2/2 = 4pa; (2)
as simultaneous equations.
Eliminating y by substituting the value of y from (1) in (2),
mV + 2mbx + b' - ipx = 0, (3)
or
mV + 2{mb - 2p)x + b^ == 0. (4)
Solving for x (see formula for quadratic. Appendix §309, (2)).
_ _ mb — 2p 2\/p' — pmb /gx
m^ ~ m^
Therefore there are in general two values of x or two points of
intersection of the straight line and the parabola. By the defini-
tion of a tangent to a curve (§146) this line becomes a tan-
gent to the parabola when the two points of intersection become
a single point; that is, when the expression under the radical in (5)
approaches zero. This condition requires that
p^ — pmb = 0,
or
b = p/m. (6)
Therefore when 6 of equation (1) has this value, the line is tangent
to the parabola. The equation of the tangent line is, therefore
y = mx + p/m. (7)
This line is tangent to the parabola y^ = ipx for all values of
§240] THE CONIC SECTIONS 419
m. Substituting in (5) the value of & = p/m, we may find the
abscissa of the point of tangency
i
xi = p/m'. (8)
Substituting this value of x in (7) the corresponding ordinate o'
this point is found to be
yi = 2p/m. (9)
240. Properties of the Parabola. In Fig. 171, F is the focus,
HK is the directrix, PT is a tangent at any point P The
perpendicular PN to the tangent at the point of tangency is
called the normal to the parabola. The projection DT of the
tangent PT on the X-axis is called the subtangent and the pro-
jection DN of the normal PN on the X-axis is called the sub-
normal. The line through any point parallel to the axis, as PR,
is known as a diameter of the parabola.
(a) The subtangent to the parabola at any point is bisected by
the vertex. It is to be proved that OT = OD for all positions of P
Now OD is the abscissa of P, which has been found to be p/m^.
From the equation of the tangent
y = mx + p/m,
the intercept OT on the X-axis is found by putting y = Q and
solving for x. This yields
X = — p/m}.
This is numerically the same as OD, hence the vertex 0 bisects
DT.
(6) The subnormal to the parabola at any point is constant and
eqval to the semi-latus rectum.
The angle DPN has its sides mutually perpendicular to the
sides of the angle DTP, hence the angles are equal. Since the
tangent of the angle DTP = m, therefore
tangent DPN = m.
From the right triangle PDN,
DN = PD tan DPN = PD m
= i2p/m)m = 2p.
420 ELEMENTARY MATHEMATICAL ANALYSIS [§241
Since KF also equals 2p, we have
DN = KF.
(fi) PFTH is a rhombus. To prove the figure PFTH a rhombus
it is merely necessary to show that FT = PH, since PF = PH
FT = FO + OT
PH = DK = DO + OK
But
OD = OT and OK = FO.
Therefore
FT = PH,
and 1;he figure is a rhombus.
It follows that the two diagonals of the rhombus intersect at
right angles on the F-axis.
(d) The normal to a parabola bisects the angle between the focal
radius and the diameter at the point. We are to show that
Z NPF = Z NPR.
Since FPHT is a rhombus,
Z FPT = Z TPH.
But
Z TPH = Z RPS,
being vertical angles. From the two right angles NPT and NPS
subtract the equal angles last named. Hence,
Z FPN = Z NPR.
It is because of this property of the parabola that the reflectors
of locomotive or automobile headlights are made parabolic.
The rays from" a source of light at F are reflected in lines parallel
to the axis, so that, in the theoretical case, a beam of light is sent
out in parallel lines, or in a beam of undiminishing strength.
241. To Draw a Parabolic Arc. One of the best ways of de-
scribing a parabolic arc is by drawing a large number of tangent
lines by the principle of §240 (c). Since in Fig. 171 the tan-
gent is for all positions perpendicular to the focal line FH at
the point where the latter crosses OY, it is merely necessary to
§241]
THE CONIC SECTIONS
421
draw a large number of focal lines, as in Fig. 177, and erect
perpendiculars to them at the points where they cross the F-axis.
The equations of the tangent lines in Fig. 177 are of the form
= mx + p/m
(1)
Fig. 177. — Graphical construction of a parabolic arc "by tangents."
in which p is the constant given by the equation of the parabola,
and in which m takes on in succession a sequence of values appro-
priate to the various tangent lines of the figure. These lines are
said to constitute a family of lines and to envelop the curve to
422 ELEMENTARY MATHEMATICAL ANALYSIS [§242
which they are tangent. The curve itself is called the envelope
of the family of lines.
The curve of the supporting surface of an aeroplane as well as
the curve of the propeller blades is a parabolic arc. The curve of
the cables of a suspension bridge is also paraboUc.
Exercises
1. Write the equation of the parabola which the family y = mx
+ 7 /2m envelops.
2. Draw an arc of a parabola if p = 3 inches.
3. At what point is ^ = mx + Z/m tangent to the parabola y' =
121?
4. At what point isy = mx + ll/ira tangent to y^ = 44x?
6. Draw the family of lines y = mx + 1/m for m = 0.4, m = 0.6,
m = 0.8, m = 1, m = 2, m = 4, m — 8.
242. Tangent to the Circle. An equation of a tangent hne to
a circle can be found, as in the case of the parabola above, by
finding, the points of intersection of
y = mx + b (1)
and
x^ + y^ = a^ (2)
and then imposing the condition that the two points of intersection
shall become a single point. The value of 6 that satisfies this
condition when substituted in (1) gives the equation of the re-
quired tangent. It is easier, however, to obtain this result by the
following method. In Fig. 178 let the straight line be drawn
tangent to the circle at T. Let the slope of this line be m.
Then m = tan ONT = tan a, if a be the direction angle of the
tangent line. The intercept b of t*he line on the F-aris can be ex-
pressed in terms of a and a,
b = a sec a = ay/l -{- m'^. (3)
Hence the equation of the tangent to the circle is
y = mx + a-\/i + m^.
The double sign is written in order to include in a single equation
the two tangents of given slope m, as illustrated in the figure.
§243]
THE CONIC SECTIONS
423
Exercises
1. Find the equations of the tangents to x' + y^ = 16 making an
angle of 60° with the X-axis.
2. Find the equations of the tangents to x' + y^ = 25 making an
angle of 45° with the X-axis.
3. Find the equation of tangents to x^ + y^ = 25 parallel to y =
3x - 2.
4. Find the equation of tangents to x^ + y' = 16 perpendicular to
y = (l/2)x + 3.
5. Find the equations of the tangents to (a; — 3)^ -|- (?/ — 4)^ = 25
whose slope is 3.
FiG. 178. — The equation of a line of given slope, tangent to a given
circle.
6. Find the equation of the tangent to the circle by the method
of §239.
243. Nonnal Equation of Straight Line. The normal equation
of the straight line was obtained in polar coordinates in §71.
The equation was written
p cos {d — a) = a. (1)
In this equation (p, d) are the polar coordinates of any point on
the line, a is the distance of the line from the origin, and a is the
direction angle of a perpendicular to the line from the origin.
(See Fig. 71.) Expanding cos {6 — a) in (1) we obtain
f> cos 6 cos a -J- p sin 5 sin a = a. (2)
424 ELEMENTARY MATHEMATICAL ANALYSIS [§243
But for any value of p and d, p cos B = x and p sin 9 = y.
Hence (2) may be written in rectangular coordinates
X cos a + y sin a = a. (3)
This also is called the normal equation of the straight line.
If an equation of any line be given in the form
ax + by = c (4)
it can readily be "reduced to the normal form. Dividing this
equation through by s/a^ + b^,
" b c .,,
X + . y = . (5)
Va' + b^ Va^ + h^ VaT+h^'
Now aly/a?' + 6^ and b/-\/a^ + h^ may be regarded as the cosine
and sine, respectively, of the angle formed with the positive Z-axis
by the line joining the origin to the point (a, 6). Calling this
angle a, equation (5) may be written
X cos a + 2/ sin o; = d, (6)
which is of the form (3) above. Inasmuch as the right-hand side
of the equation in the normal form represents the distance of the
line from the origin, it is best to keep the right-hand Side of the
equation positive. The value of a and the quadrant in which it
lies is then determined by the signs of cos a and sin a on the left-
hand side of the equation. The angle a may have any value from
0° to 360°.
Illustrations:
• (1) Put the equation Zx — ^y = 10 in the normal form. Here
a2 + 62 = 3' + ( - 4)2 =25. Dividing by 5 we obtain
{Z/5)x - (4/5)2/ = 2.
The distance of this line from the origin is 2. The angle a is the angle
whose cosine is 3/5 and whose sine is — 4/5. Therefore, from the
tables, a = 306° 52'.
(2) Put the equation 3a; — 4v 4- 20 = 0 in the normal form. Trans-
posing and dividing by — 1 to make the right-hand side of the equa-
tion positive, we obtain — 3a; -(- 4i/ = 20.
Here cos a = - 3/5, sin a = 4/5, d = 4. Hence a = 126° 52'.
§244] THE CONIC SECTIONS 425
(3) What is the distance between the lines (1) and (2)? The lines
are parallel and on opposite sides of the origin. Their distance apart
is therefore 2 + 4 or 6.
(4) Put X + y = I in the normal form. Here VoM- b^ = V'2.
The equation becomes i -s/^x + i ■\/2y = i y/2. a = 45°, d = i \/2.
Exercises
1. The shortest distance from the origin to a line is 5 and the direc-
tion angle of the perpendicular from the origin to the line is 30°-
Write the equation of the line.
2. The perpendicular from the origin upon a straight line makes an
angle of 135° with OX, and its length is 2v'2. Find the equation of
the line.
3. Write the equation of a straight line in the normal form if a =
60° and d = y/s.
4. Put 2\/3a; + 2?/ = 32 in the normal form and find the
numerical values of a and d.
6. Put 2a; — 2i/ = 1 in the normal form and find the values of a
and d.
6. Find the equation of the straight line, if the perpendicular from
the origin on the line, makes an angle of 46° with the X-axis and its
length is ■\/2.
7. Put 2 -ho = 1 ii the normal form.
244. To Translate Any Locus a Given Distance in a Given Direc-
tion. To move any locus the distance d to the right we sub-
stitute (xi — d) for X in the equation of the locus. To move the
locus the distance d in the y direction we substitute (2/1 — d) for y.
To move any locus the distance d in the direction a we substitute
(xi — d cos a) for x, , ,
(2/1 — d sin a) for y,
which must give the desired equation of the new locus. It is
not necessary to use the subscript attached to the new coordinates
if the distinction between the new and old coordinates can be
kept in mind without this device.
The circle x^ + y^ = a^ moved the distance d in the direction
a becomes
(x — d cos a)^ -i- (y — d sin a)" =• a'^
426 ELEMENTARY MATHEMATICAL ANALYSIS [§245
which may be changed to
x^ — 2Ax, cos a + 2/^ — 2^2/ sin a = a?- — cP.
245. Distance of Any Point From Any Line. Let the equation
of the line I, Kg. 179, in the normal form be
X cos a + 2/ sin a = a, (1)
and let (a;i, t/i) be any point P in the plane. (See Fig. 179.)
If the point {xi, yi) is on the opposite side of the line froin the
origin, the line can be moved so that it will pass through the point
by translating it the proper distance in the direction a. Let
the unknown amount of the required translation be represented
by d. To translate the line the amount d in the direction a,
we must substitute for x and y the values
X = x' — d cos a . .
y = y' — d cos a
We obtain
(x' — d cos a) COB a + iy' — d sin a) sin a =a. (3)
The line represented by this equation passes through the point
(xi, 2/i). Substituting these coordinates for x' and y' and solving
for d, we have
d = Xi COS a + yi sin a — a. (4)
This is the distance of (xi, yi) from the given line.
If the given point is on the same side of the line as the origin,
as the point Pixi, y^ Fig. 179, then the given line must be
translated the distance d in the direction (180° + a), and the result
is the same as (4) above except all signs are changed. We are usu-
ally interested only in the numerical value of d, so that formula
(4) may be used for all cases. When the value of d comes out
negative it merely means that the given point is on the same side
of the line as the origin.
Equation (4) may be interpreted as follows :
To find the distance of any point from a given line, put the equa-
tion of the line in the normal form, transpose all terms to the left-
hand member and siibstitvte the coordinates of the given point for x
and y. The absolute value of the left-hand member is the distance
of P from the line.
§245]
THE CONIC SECTIONS
427
The above facts may be stated in an interesting form as follows:
Let any line be
X cos a + 2/ sin a — a = 0.
If the coordinates of any point on this line be substituted in this
equation, the left-hand member reduces to zero. If the coordi-
nates of any point not on the line be substituted for x and y in the
equation, the left-hand member of the equation does not reduce
to zero, but becomes negative if the given point is on the origin
side of the line and positive if the given point is on the non-
origin side of the line . The absolute value of the left-hand member
\
\'^
Y
\
\
\
2
\
\
\
\
\
cZ^APi(<Kny,)
\
\
\
\
4^\
\
k\ \
\
.
\,-y\ ^
\
>
^
a \ \
P.ik^,
!/=) > \
\
<^ \ \
0
/-
/
\ \ -A \
\ \ \_
Fig. 179. — Distance of any point from a given Une.
in each case gives the distance of the given point from the line.
Thus every line may be said to have a "positive side" and a
"negative side." The "negative side" is the side toward the
origin.
Illtjstbation 1. Find the distance of (—1, 4) from the line 3a; —
41/ = 10.
Transpose and put left-hand member in normal form
1^ — iV
0.
428 ELEMENTARY MATHEMATICAL ANALYSIS [§246
Subatitute — 1 for a; and 4 for y. The left member is now the value
of d, BO that
The result is negative, so that ( — 1, 4) is on the same side of the line as
the origin.
iLLtrsTRATioN 2. Find the distance of (2, — 4) from
x-2 y+3
4 7
Clear of fractions and simplify,
7a; - 4?/ - 26 = 0.
Put in normal form,
iz^ - i\y - u = 0.
Substitute 2 for x and — 4 for y,
The point is on the non-origin side of the given line, and irV of one imit
from it.
Exercises
1. Find the distance of the point (4, 5) from the line 3x + iy = 10.
2. Find the distance from the origin to the line x/3 — 2//4 = 1.
3. Find the distance from (-3, - 4) to 12(a; + 6) = 6(2/ - 2).
4. Find the distance from (3, 4) to the line x/3 — y/4 = 1.
5. Find the distance between the parallel lines y = 2x -\- 3, and
y = 2x +5.
6. Find the distance between y = 2x — 3, y = 2x + 5.
7. Find the distance from (0, 3) to 4a; — 3y = 12.
8. Find the distance from (0, 1) to a; + 2 — 2?/ =0.
246. Tangent to a Circle at a Given Point. The equation of
a line having a given slope m and tangent to a given circle with
center at the origin, was given in §242. We shall now find the
equation of the line that is tangent to the circle at a given point
(xo, 2/o).
The line,
a ='p cos {6 — a), (1)
or its equivalent,
a; cos a + 2/ sin a = a, (2)
§247]
THE CONIC SECTIONS
429
is tangent to the circle of radius a, and the point of tangency is at
the end of the diameter whose direction angle is a. The point of
tangency is therefore (o cos a, a sin a). Hence, multiplying (2)
through by a, we obtain
x{a cos a) + y(a sin a) = a^, (3)
or
xox + yoy = a.K (4)
This is the equation of the hne tangent to the circle of radius a
at the point (xq, ya).
Thus 3a; + 4?/ = 25 is tangent to a;^ + y"^- = 25 at (3, 4).
247. Tangent to the Ellipse at a Given Point. It is easy to
draw the tangent to the ellipse at any desired point. In Fig. 180,
Fig. 180. — Tangent to the ellipse at a given point.
let Po be the point at which a tangent is desired. Draw the
major circle, and let Pj of the circle be a point on the same ordinate
asPo. Draw a tangent to the circle at Pi and let it meet the
Z-axis at T. Then when the circle is projected to form the
ellipse, the straight line PiT is projected to make the tangent to
the ellipse. Since T when projected remains the same point and
since Po is the projection of Pi, the line through Po and T is the
required tangent to the ellipse.
430 ELEMENTARY MATHEMATICAL ANALYSIS [§248
The equation of the tangent PoT is alao readily found. The
equation of PiT is
xxo + yyo = ffl^ (1)
To project his into the line Po^ it is merely necessary to multiply
the ordinates y and yo' by b/a; that is, to substitute y = ay/b and
2/o' = ayo/b. Whence (1) becomes
xox + a^yoy/b^ = a^ (2)
or, dividing by a^,
xox/a" + joj/b^ = I (3)
which is the tangent to
a;2/a2 -I- 2^2/62 = 1
at the point (xo, yo)-
Exercises
1. Find the equations of the tangents to the ellipse whose semi-axes
are 4 and 3 at the points for which x = 2.
2. Find the equations of the tangents to x'/16 + y^/9 = 1 at the
ends of the left-hand latus rectum.
3. Required the tangents to x'/9 + y'/i = 1, making an angle of
45° with the X-axis. [Solve y = x + b and x'/9 + y"/4 = 1 as in
§239.]
4. Find the equatioiis of the tangents to sr^/lOO -I- y'/25 = 1 at
the points where y = 3.
6. Find the equations of the tangents to x'/S6 + y'/16 = 1 at the
^ints where x = y.
248. The Tangent, Normal, and Focal Radii of the Ellipse. In
the right triangle PiOT, Fig. 180, the side PiO is a mean propor-
tional between the entire hjrpotenuse OT and the adjacent
segment OD. That is
a^ = xoX OT, or OT = a^/xo
Then FiT = OT - OFi = OT - ae
= a'^/xo — ae
Wkewis^ FtT = OT + OFi
5= aVaio ■+- ae
§249] THE CONIC SECTIONS 431
Therefore FrT/FiT = {a^xa - ae)/{ayxa + ae)
= (a — exo)/{a + exo)
But by §230 this last ratio is equal to r^/r^. Therefore we
may write FiT/F^T = P^F./PoF,.
Hence T, which divides the base FiFi of the triangle PoFaFi
externally at T in the ratio of the two sides PF^ and PFi of the
triangle, lies on the bisector of the external angle FiPoQ of the
triangle FJPoFi. This' proves the important theorem:
The tangent to the ellipse bisects the external angle between the
focal radii at the point.
This theorem provides a second method of constructing a
tangent at a given point of an elUpse, often more convenient
than that of §247, since the method of §247 often runs the
construction off of the paper.
The normal PoN, being perpendicular to the tangent, must
bisect the internal angle F^PoFi between the focal radii F^o and
F,Po.
Since the angle of reflection equals the angle of incidence for
light, sound, and other wave motions, a source of light or sound at
Fi is "brought to a focus" again atF^, because of the fact that the
normal to the ellipse bisects the angle between the focal radii.
249. Additional Equations of the Straight Liae.^ The equations
of the straight line in the slope form
y = mx + b (1)
and in the normal forms
p cos {d— a) = a (2)
a; cos a + 2/ sin a = a (3)
and the general form
ax + by + c = 0 (4)
have already been used. Two constants and only two are neces-
sary for each of these equations. The constants in the first
equation are m and 6; in the second and third, a and a; in the
fourth a/c and b/c, or any two of the ratios that result from divid-
ing through by one of the coefficients. Equation (4) appears to
contain three constants, but it is only the relative size of these that
> See §17.
432 ELEMENTARY MATHEMATICAL ANALYSIS [§249
determines the particular line represented by the equation, since
the line would remain the same when the equation is multiplied
or divided through by any constant (not zero).
These facts are usually summarized by the statement that two
conditions are necessary and sufficient to determine a straight
line. The number of ways in which these conditions may be given
is, of course, unhmited. Thus a straight line is determined if we
say, for example, that the line passes through the vertex of an
angle and bisects that angle, or if we say that the line passes
through the center of a circle and is parallel to another line, or if
we say that the straight Une is tangent to two given circles, etc.
An important case is that in which the line is determined by the
requirement that it pass through a given point in a given direc-
tion. The equation of the line adapted to this case is readily
found. Let the given point be {xi, yi). The line through the
origin with the required slope is
y = mx.
Translate this line so that it passes through (xi, yi) and we have
y - yi = m(x - xi). (5)
Another way of obtaining the same result is : Substitute the
coordinates (.Xi, yi) in (1)
2/1 = mxi + 6. (6)
Subtract the members of this from (1) above, so as to eliminate
b. There results
y - yi = m{x — a;i). (7)
This is the required equation. The given point is (a;i, 2/1) and the
direction of the line through that point is given by the slope m.
Another important case is that in which the straight line is
determined by requiring it to pass through two given points.
Let the second of the given points be (012, yi). Substitute these
coordinates in (5)
2/2 - 2/1 = »»(»2 - a;i). (8)
To eliminate ?n, divide the members of (7) by the members of (8)
2/ - 2/1 X — xi
2/2 - 2/1 2:2 - X\
(9)
§250] THE CONIC SECTIONS 433
or, as it is usually written
L^Zli ^Yl^zll^; (10)
X — Xi X2 — Xi
0
This is the equation of a line passing through two given points.
Since (10) may be looked upon as a proportion, the equation may
be written in a variety of forms.
250. The Circle Through Three Given Points. In general, the
equation of a circle can be found (when three pojnts are given.
Either of the general equations of the circle
{x - hY +{y - kY = a\ (1)
or
a;" + 2/^ + 2gx + 2/2/ + c = 0 (2)
contains three unknown constants, so that in general three con-
ditions may be imposed upon them. It is best to illustrate the
general method by a particular example. Let the three given
points be (— 1, 3), (0, 2), and (5, 0). Then since the coordinates
of these points must satisfy the equation of the circle, we obtain
from (2) above
1 + 9 - 2ff + 6/ + c = 0, (3)
4 + 4/ + c = 0, (4)
25 +\0g + c = 0. (5)
Eliminating c from (3) and (4) and from (4) and (5), we obtain
6 - 2sr^ + 2/ = 0,
21 + 10^ - 4/ = 0.
Eliminating /
? = - 5i
Whence
/ = - 8i
and
a = 30.
So the equation of the circle is ■
a;2 4- yi _ lla; _ ny + 30 = 0.
28
434 ELEMENTARY MATHEMATICAL ANALYSIS [§251
Exercises
1. Knd the equation of the line passing through (2, 3) with slope
2/3.
2. Find the equation of the line passing through (2, 3), and (3, 5).
3. Find the line passing through (2, — 1) making an angle whose
tangent is 2 with the Z-axis.
4. Find the line through (2, 3) parallel to 2/ = 7a; + 11.
5. A line passes through (—1, — 3) and is perpendicular to
y — 2x —' 3. Find its equation.
6. Find the line passing through (— 2, 3), and (—3,-1).
7. Find the equation of the line which passes through (— 1, —3),
and (-2,4).
8. Find the slope of the line that passes through ( — 1, 6 ), and
(-2,8).
9. Find the equation of the line passing through the left focus and
the upper end of the right latus rectum of a;2/2S + y'/9 = 1.
10. Find the equation of the circle passing through (2, 8)„ (5, 7),
and (6, 6) .
11. Find the equation of the circle which passes through (1, 2),
(- 2, 3), and (- 1, - 1).
12. Find the equation of the parabola in the form y^ = ipx which
passes through the point (2, 4).
251. Change from Polar to Rectangular Coordinates. The
relations between x, y of the Cartesian system and p, 6 of the
polar system have already been explained and use made of
them. The relations are here brought together for reference :
X = p cos 8 (1)
y = p sin 0. (2)
By these we may pass from the Cartesian equation of any locus
to the equivalent polar equation of that locus. Dividing (2)
by (1) and also squaring and adding, we obtain:
e = tan-i y/x (3)
P = Vx^ + y^ (4)
These may be used to convert any polar equation into the Carte-
sian equivalent.
262. Rotation of Any Locus. It has already been explained
that any locus can be rotated tlirough an angle a by substituting
§252]
THE CONIC SECTIONS
435
(Ot — a) for d in the polar equation of the locus. It remains to
determine the substitutions for x and y which will bring about
the rotation of a locus in rectangular coordinates. Let us consider
any point P of a locus before and after rotation through the given
angle a. Call the coordinates of the point before rotation
(x, y) in rectangular coordinates and (p, 6) in polar coordinates.'
Then, from (1) and (2), §251,
X = p cos 6 (1)
t/ = P sin 6. (2)
Call the coordinates of the point after rotation (xi, yi) and
(Pi, ^i), but note that the value of p is
unchanged by the rotation. Then for p (Pi, e,) or
the point P', Fig. 181, we may write ] /\ (X1.V1)
Xi = p cos 61 (3)
2/1 = p sin 01. (4)
Since, however, the rotation requires —
that Fig- 181
6 = Oi - ot (5)
equations (1) and (2) become
X = p cos (di -^ a) = p cos di cos a + p sin 81 sin a (6)
2/ = p sin (di — a) = p sin 61 cos a — p cos 0i sin a. (7)
But, from (3) and (4), p cos di and p sin di are the new values of
X and y; hence, substituting in (6) and (7) from (3) and (4) we
obtain
X = xi cos Qi + yi sin a (8)
y '= yi cos a — Xi sin a. (9)
Hence, if the equatiofi of any locus is given in rectangular co-
ordinates, it is rotated through the positive angle a by the sub-
stitutions
X cos a + y sin a for X
y cos Q! — X sin a for y, (10)
in which it is permissible to drop the subscripts, if the context
shows in each case whether we are^ dealing with the old x and y
or with the new x and y.
PiP.e) or
-Rotation of
any locus.
436 ELEMENTARY MATHEMATICAL ANALYSIS [§252
If the required rotation is clockwise, or negative, we must
replace a by (— a) in aU of the above equations.
Whenever comenient, the eqvation of a curve should he taken in
the -polar form if it is required to rotate the locus.
Important Facts: The following facts should be remembered
by the student:
(1) To rotate a curve through 90°, change x to y and y to { — x).
(2) Rotation through any angle leaves the expression x^ + y^
(fir any function of it) unchanged. This is obvious since the circle
x^ + y^ = a^ is not changed by rotation about (0, 0).
Fig. 182. — Effect of rotation on the special forms x^ + y^, 2xy, and
x^ - 2/2.
(3) Rotation through + 45° changes 2xy to y^ — x^.
Rotation through — 45° changes 2xy to x^ ,— y^.
(4) Rotation through + 45° changes x^ — y^ to 2xy.
Rotation through — 45° changes x'^ — y^ to — 2xy.
Statements (3) and (4) follow at once from consideration of the
equations
i;2 - 2/2 = a^ (1)
a' (2)
a' <3)
a^ (4)
2xy
yi _ j;Z
- 2xy
§253] THE CONIC SECTIONS 437
of the four hyperbolas bearing corresponding numbers (I), (2),
(3), (4) in Fig. 182. The proper substitution in any case can
be remembered by thinking of the four hyperbolas of this figure.
(5) The degree of an equation of a locus cannot he changed by
a rotation. This follows at once from the fact that the equations
of transformation (8) and (9) are linear.
Exercises
In order to shorten, the work, use statements (1) to (4) whenever
possible.
1. Turn the locus a* — j/' = 4 through 45°.
2. Turn x' + y' = a' through 79°. Turn ixy = 1 through 45°.
3. Turn x cos a -\- y sin a = a through an angle /3. (Since this
locus is well known in the polar form, transformation formulas (6) and
(7) above need not be used.)
4. Rotate x' - y^ = 1 through 90°.
5. Rotate s" — j/^ = a' through — 45°.
6. Rotate x' - y' = 1 through 30°.
7. Rotate x^ - y" = 4: through 60°.
8. Rotate ixy = 1 through 30°.
9. Rotate x' + 2y' = 1 through 45°.
10. Change the equation (x — a)' + (y — b)' = rHo the polar form.
11. Change p cos 29 = 2a, one of a class of curves known as Cote's
spirals, to the Cartesian form.
12. Write the equation of the lemniscate in the polar form.
13. Show that p' — 2p/oicos (9 — 9i) + pi" = a'is the polar equation
of a circle with center at (pi, 9i) and of radius a.
14. Write the Cartesian equation of the locus p" = 16 sin 29.
15. Turn p^ = 8 sin 29 through an angle of 45°.
16. Rotate x^ - 2y^ = 1 through 90°.
17. Rotate {x^ + y^)^ + {x' - y')^ = 1 through 45°.
18. Rotate log {x' + y^) = tan {x^ - y^) through 45°.
19. Rotate x^ -&xy -\-y''=\ through 45°.
20. Rotate x^ + y^^ = a}^ through 45°. Show that the result is
the parabola y = x^la + o/2, and sketch the curves.
253. Ellipse with Major Axis at 45° to the QX Axis. The
ellipse frequently arises in applied science as the resultant of the.
projection of the motion of two points moving uniformly on two
438 ELEMENTARY MATHEMATICAL ANALYSIS [§253
circles, as has already been explained in §186. Thus the para-
metric equations «
X = a cos t (1)
2/ = 6 sin t,
(2)
define an ellipse which may be considered the resultant of two
S.H.M. in quadrature. We shall prove that the equations
a; = o cos t
y = asm (t + a),
(3)
(4)
define an ellipse, with major axis making an angle of 45° with OX.
The graph is readily constructed as in Fig. 183. The Car-
tesian equation of the curve is found by eliminating t between
/
/
^
Y
B
|\
Ic
—
■P'
\^ "^
y
¥ \y
/
^'
A
X'
/
.^^
■^
1 y\- \x
/
,
-^
O
/
A
T 3
.■'•
/
1
-^
y
/
c
^
^
r'
Fig. 183. — The ellipse x = a cos t,y — a sin (< + a).
(3) and (4). Expanding the sin {t + «) in (4) and substituting
from (3) we obtain
(5)
y = X sin.a + -y/a^ — x"^ cos a.
Transposing and squarimg
a;^ — 2xy sin a + j/'' = a^ cos' a. (6)
By §252 rotate the curve through an angle of (— 46''). We
§254] THE CONIC SECTIONS 439
know that (x* + y^) is unchanged and that 2xy is to be replaced
by (a;2 — y^). Therefore (6) becomes
x^l — sin a) + y^{l + sin a) = a^ cos^a. (7)
Replacing cos^ a by 1 — sin^ a, and dividing through by the
right-hand member, we obtain
a\l + sin a) "^ a^l - sin a) " ^' ^^^
which may be written
—-— + -^,= 1, (9)
2a2 cos^ I 20^ sin'' |^
where /8 is the complement of a. Equation (8) or (9) proves
that the locus is an ellipse. It is any ellipse, since by properly
choosing a and a the denominators in (8) can be given any desired
values. Hence the pair of parametric equations (3) and (4), or
the Cartesian equation (5) represents an ellipse with its major axis
inclined + 45° to the X-axis.
254. She.ar of the Circle. The effect of the addition of the term
mx to f(x), in the equation y = f(x), has been shown in §38
to change the shape of the locus by lamellar, or shearing, motion
in the Xy-plane. We usually speak of this process as "the shear of
the locus y = fix) in the line y — mx." When appUed to the circle
?/ = + V'a^ — x'^ the effect is to move vertically the middle point
of each double ordinate of the circle to a position on the line
y = mx. The result of the shearing motion is shown in Fig. 184.
The area hounded by the curve is unchanged by the shear.
The equation after shear is •
y = mx + ■\/a'^ — x^. (1)
This is the same form as equation (5) of §253, if we put m =
and replace y cos ahyyu After the substitution, rotate the curve
sma
cos a
440 ELEMENTARY MATHEMATICAL ANALYSIS [§255
through 45°, and replace ^i by y cos a. The equation can then
be written
a^ (1 + sin a)
y^ (1 + sin a)
(2)
Therefore the curve of Fig. 184 is an ellipse.
The straight line y = mx passes through the middle points of
the parallel vertical chords of the eUipse
y = mx +
(3)
The locus of the middle points of parallel chords of any curve is
called a diameter of that curve. We have thus shown that one
diameter of the eUipse is a straight line. Since the same reasoning
applies to
y = mx+ {h/aWa^ - x\ (4)
X' /
4 .
0 / A
w
Fig. 184. — The ellipse looked upon as the shear of the circle OA in a
Une M'OM.
which may be regarded as any ellipse in any way oriented with
respect to the origin, the proof shows that the mid-points of arbi-
trarily selected parallel chords of an eUipse is always a straight
line.
266. General Equation of the Second Degree. The general
equation of the second degree in two variables may be written in
the standard form
oa;2 + 2hxy + &2/' + "igx + 2/3/ + c = 0.
(1)
§256]
THE CONIC SECTIONS
441
In treatises on Conic Sections it is shown that the general
equation of the second degree in two variables represents a conic.
Three cases are distinguished as follows:
The general equation of the second degree represents
an ' eUipse if h^ — ab < 0
a parabola ii h^ — ab = 0
a h3rperboIa if h^ — ab > 0.
(2)
(3)
(4)
To render the above classification true in all cases we must classify
the "imaginary ellipse," -^ +r^ = — 1, as an ellipse, and other
degenerate cases must be similarly treated. The expression
h^ — ab is called the quadratic invariant of the equation (1), so
Fig. 185. — Confocal ellipses and hyperbolas. Note that the curves of
, one set cut the curves of the other set orthogonally.
called because its value remains unchanged as the curve is moved
about in the coordinate plane. In other words, as the locus (1)
is translated or rotated to any new position in the plane, and while
of course the coefficients of x^, xy, and y^ change to new values, the
function of these coefiicients, h'^ — ab, does not change in value,
but remains invariant. The above facts are not proved in this
book.
442 ELEMENTARY MATHEMATICAL ANALYSIS [§256
266. Confocal Conies. Fig. 185 shows a number of ellipises
and hyperbolas possessing the same foci A and B. This family
of curves may be represented by the single equation,
in which the parameter k takes on any value less than a^, and in
which a > b. If fc satisfies the inequality
k < 6^
the curves are ellipses. If k satisfies the inequalities
¥ < k < a\
the curves are hyperbolas. The ellipses of Fig. 185 may be
regarded as representing the successive positions of the wave front
of sound waves leaving the sounding body AB, or they may be
regarded as the equipotential lines around the magnet AB, of
which the hyperbolas represent the lines of magnetic force.
Exercises
1. Sketch the curve y = 2x + \/4 — as".
2. Draw the curve
X = 2 cos B
2/ = 2 sin (9 +7r/6).
3. Find the axes of the elUpse
X = 3 cos 6
2/ = 3 sin (9 + 7r/4).
4. Draw the curve y = x + \/6x — x*.
6. Draw the curve y = x + y/x' — 6i.
6. Show that y = x ± \/6x is a parabola.
7. Sketch the curve y = (I/2)x + Vl6 - x^
8. Sketch the curve 2/ = 5x sin 60° + cos 60°-\/25 -x«.
9. Discuss the curve
x'/a' + y'/b" - 2{xy/ah) sin a = cos^ a.
§256] THE CONIC SECTIONS 443
Show that the locus is always tangent to the rectangle x = ± o,
y = ± b, and that the points of contact form a parallelogram of
constant perimeter 4:\/a' + b^ for all values of a. Hint: Compare
with equation (6), §253.
10. Show that x = a cos {8 — a), y = b sin (9 — or) represents an
elUpse for all values of a. ■
11. Prove from equation (8), §253, that the distance from the
end of the minor to the end. of the major axis of the resulting ellipse
remains the same independently of the magnitude of a.
12. Show that the following construction of the hyperbola xy' = a'
is correct. On the — X-axis lay off OC = a. Connect C with any
point A on the F-axis. At C construct a perpendicular to AC cut-
ting the F-axis in B. At B erect a perpendicular to BC cutting the
-|- X-axis at D. Through A draw a parallel to the X-axis and through
D draw a parallel to the F-axis. The two lines last drawn meet at P,
a point on the desired curve.
13. Explain the following construction of the cubical parabola
a^y = x'. Lay off OB on the — F-axis equal to a. From B draw a
line to any point C of the X-axis. At C erect a perpendicular to BC
cutting the F-axis at D. At D erect a perpendicular to CD cutting
the X-axis at E. Lay off OE on the F-axis. Then OE is the ordinate
of a point of the curve for which the abscissa is OC.
14. Explain and prove the following construction of the semi-
cubical parabola, ay^ = i'. Lay off on the — X-axis, OA = u..
From A draw a parallel to the line y = mx, cutting the F-axis in B.
Erect at B a perpendicular to AB cutting the X-a,xis at C, and at C
erect a perpendicular to Ou. The point- of intersection with y = mx
is a point of the curve.
Miscellaneous Exercises
1. Show that sec^ a(l + sec 2a) = 2 sec 2a.
sin a + sin 2a
2.
Show that ;; — j 1 s — = tan a.
1 "1- cos a -\- cos 2a
3.
oL ii. J. COS a -|- sin a cos a — sin a
Show that -. i — -. — = 2 tan 2a.
cos a — sm. a cos a -|- sm a
4.
„, ^, ^ cos (a — ff) 1 + tan a tan fi
bnow that 7 — ;— ts — 'i 1 1 ;;'
cos (a -f- (3) 1 — tan a tan /3
. 1 -1- tan" 1
Show that . = sec a.
5.
1-tan-^
444 ELEMENTARY MATHEMATICAL ANALYSIS [§256
6. Solve the equation sin* a — 2 cos a + i =0.
7. Simplify the product
(x -2 - v'3)(a; - 2 - iVs){x - 2 + V3)ix -2 + iVS).
8. Express in the form c cos (o — 6) the binomial
30 cos o + 40 sin a.
9. Find tan 6 by means of the formula for tan (A + B), if
8 =tan-i 1/2 + tan-i 1/3.
10. Find sin 9, if 9 = sin-i 1/5 + sin-i 1/7.
11. Find the "equation of a circle whose center is the origin and
which passes through the point (14, 17).
12. Discuss the curve
X = aS
y = a(l — cos 0).
• 13. Graph on polar paper p^ = a' cos 29.
14. A fixed point located on one leg of a carpenter's "square"
traces a curve as the square is moved, the two arms of the square,
however, always passing through two fixed points A and B. Find
the equation of the curve.
16. Find the parametric equations of the oval traced by a point
attached to the connecting rod of a steam engine.
16. Prove that
tan (45° + t) - tan (45° - r) = . ^^f'^! ■
17. Find the quotient of (6 - 2i) by (3 + 75i).
18. Solye for y by inspection:
sin (90° + iy) cos (90° - iy) + cos (90° + ^y) sin (90° - iy) = sin y.
19. Write the parametric equations for the circle, the ellipse, and
the hyperbola.
20. The length of the shadow cast by a tower varies inversely as
the tangent of the angle of elevation of the sun. Graph the length
of the shadow for various elevations of the sun.
21. From your knowledge of the equations of the straight line and
circle, graph y = ax + y/a^ — x'^-
(See Shearing Motion, §37.)
22. In the same manner, sketch y = a -{■ x -\- y/'a^ — x^-
23. Graph the curve y = 1/x + x'. Has this, curve a minimum
point?
§256]
THE CONIC SECTIONS
445
24. Find by use of logarithmio paper the equations of the curves
of Fig. 186. These curves give the amounts in ce^ts per kilowatt-
hour that must be added to price of electric power to meet fixed
charges of certain given annual amounts for various load factors.
25. The angle of elevation of a mountain top seen from a certain
point is 29° 4'. The angle of depression of the image of the mountain
top seen in a lake 230 feet below the observer is 31° 20'. Find the
height and horizontal distance of the mountain top, and produce a
single formula for the solution of the problem.
26. Find the points of intersection of the curves
a;2 -I- 2/^ = 4
y^ = 4x.
27. Solve llOx-* + 1 = 2\x-'.
28. Solve 3(a; - 7)(a; - l){x - 2) = (» + 2){x - 7){.x + 3).
29. Solve sin x cos x = 1/4.
30. By means of a progression, show how to find the compound
interest on $1000 for 25 years at 5 per cent.
100
90
80
70
I
I 50
•s
o
^ 40
\
-r
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
s
S
s
^'
^•>
\
s
\
s
s
^
■fe,
s
■v
^
^
^
-
.^
!a
'■a
1 ]
^rn
*-.
~
^
^
J
is.
fSc
1
hi
Vr,
as
~
~
-
~
"■
—
0.1 0.2 0.3 0.4 05
0.6 0.7 0.8 0.9 1
Cents per K.W.Hour
11 1.2 1.3 1.4 15
Flo. 186.— Annual fixed charges of $10, $15, and $20 of a certain hydro-
electric plant, reduced to cents per kw-hr for various load factors.
446 ELEMENTARY MATHEMATICAL ANALYSIS [§256
31. Find the approximate equations for the following data:
(a) Steam pressure :
(6) Gas-engine mixture :
V = volume, p = pressure.
V = volume, -p = pressure.
(a)
(6)
r
V
V
V
2
68.7
3.54
141.3
4
31.3
4.13
115.0
6
19.8
4.73
95.0
8
14.3
5.35
81.4
10
11.3
5.94
6.55
7.14
71.2
63.5
54.6
32. Show that p^ = a^ cos 29 is the polar equation of a lemniscate,
33. When an electric current is cut off, the rate of decrease in the
current is proportional to the current. If the current is 36.7 amperes
when cut off and decreases to 1 ampere in one-tenth of a second,
determine the relation between the current C and the time t.-
34. Write four other equations for the circle p = 2-\/3 sin 9 —
2 cos e.
36. Write four other equations for the sinusoid j/ = sin x —
■\/3 cos X.
36. Find the angle that Zx + iy = 12 makes with ix - Zy = 12.
37. From the equation
9 = 6 sin (2« - 1°)
determine the amplitude, period, and frequency of the S.H.M.
38. A simple sinusoidal wave has a height of 3 feet, a length of 29
feet, and a velocity of 7 feet per minute. Another wave with the
same height, length, and velocity lags 15 feet behind it. Give the
equation of each.
39. Simplfy
(3\/3 - 3i) 2( - 1 -I- VZiy (cos 36° + i sin 36°) (cos 20° + i sin 20°) .
(2 -I- 2^31)
2(cosll° +isin 11°)
§256] THE CONIC SECTIONS 447
40. Calculate (1 - VsifK
41. Plot the amount of tin required to make a tomato can to hold
1 quart as a function of the radius of its base. Deterinine approxi-
mately from the graph the dimensions requiring the least tin.
42. Find the axes of the ellipse whose foci are (2, 0) and ( — 2, 0),
and whose directrices are x = ± 5.
43. Write the polar equation for the ellipse in problem 42.
44. Find the equation of the hyperbola whose foci are (5, 0) and
(— 5, 0), and whose directrices are x = + 2.
46. Write the equation of the hyperbola of problem 44 in polar
coordinates.
46. Discuss the curve p(l + cos 9) = 4. Write its equation in
rectangular coordinates.
47. Find the foci of the hyperbola 2xy = a". Also its eccentricity.
48. Find the equation of the locus of a point whose distance from
the point (3, 4) is always twice its distance from the line 3x + 4y = 12.
What is the locus?
49. A point moves so that the quotient of its distance from two
fixed points is a constant. Find the equation of the locus of the
point.
60. Evaluate log 10 - log2 8 + logy 492.
51. Find the maximum and minimum value of (3 sin j: — 4 cos x).
What values of x give these maximum and minimum values?
62. Find the equation of a circle passing through the points (1, 2),
(- 1, 3), and (3, - 2).
63. A sinusoidal wave has a wave-length of ?r, a period of tt, and an
amplitude of t. Write its equation.
64. Compute the value of each of the following :
1^; 7 ois 47° X 6 cis (- 14°); (7 +61)"; '^Ti+ST.
65. Prove by the addition formulas that:
sin (90° -t) = COST, sin (360° - t) = - sinr,
sin (90° +t) = COST, tan (r + 270°) = - cotT.
56. Solve x2 + 6x + Vx' -|- 6x -|- 1 = 1-
57. Find the product of 3 - 2i by - 2 + i.
68. Find all the values of
(cos e -I- 1 sin e)2; (cos S + i sin 6)^^; -^V, VT.
448 ELEMENTARY MATHEMATICAL ANALYSIS [§250
69. Show that
sin (a + 6 + c) = sin o cos b cos c + cos a sin b cos c
+ cos a cos b sin c — sin o sin 6 sin r
60. Draw upon squared paper, using 2 cm. = 1, the curve y" = x
By counting the small squares of the paper find the area bounded by
the curve and the ordinates x = 1/2, 1, IJ, 2, 2i, 3, 3i, 4, . .By
plotting these points upon some form of coordinate paper, find the
functional relation existing between the x coordinate and the area
imder the curve.
61. The latitude of two towns is 27° 31'. They are 7 miles apart
measured on the parallel of latitude. Find their difference in
longitude.
62. Solve 3"'"' = 2'+'. Be very careful to take account of all
questionable operations. There are two solutions.
63. Find (two problems) the equation connecting:
X
y
6.8
19.0
14.2
21.6
21.8
23.2
32.0
26.3
46.5 .
31.5
65.0
39.1
78.0
47.0
X
y
1.3
21
2.0
25
2.8
29
3.7
33
4.3
35
5.3
38
64. Find the wave length, period, frequency, ampUtude, and velocity
for y = 10 sin (2x - 30.
66. Prove that
csc^ A „ .
— 5— j ^ = sec 2A.
csc^ A — 2
66. Find the equation of the elhpse, center at the origin, axes coin-
ciding with coordinate axes, passing through the point ( — 3, 5) and
having eccentricity 3/5.
, 67. Prove (esc 2x) (1 — cos 2x) = sin x sec x.
(esc X) (1 — cos x) = ?
68. A S.H.M. has amplitude 6, period 3. Write its equation if
time be measured from the negative end of the oscillation. State the
difference between a S.H.M. and a wave.
§250] THE CONIC SECTIONS 449
69. Sketch on squared paiper :
y =
V
y =
2'
y = logz X
y =
3"
y = logs X
y =
5^
y = logs X
y =
10-
y = logio X
70. Solve 3» - 2.T = 1.
71. Sketch
p = a, p = sec 6,
p
=
a sin e,
1
p = -, p = a cos 9,
p
=
— a cos 9,
p = (2 — cos e),
p
=
2 cos 6/ - 3,
p = — o sin 9,
P = a — a cos 8,
P = cos 9 + sin 6.
72. Simplify the expression
sin {^ - rj sec ^1 +j;j - sin Q + r^ sec (| - rj
73. Simplify and represent graphically
y«-)\^'^^) (1 +»)a + 2i).
74. Find the coordinates of the center, the eccentricity, and
the lengths of the semi-axes of: (o) rc^ + Sx + j/^ = 7, (6) x^ -\- 2x
+ 42/2 - 32/ = 0, (c) a;2 - a; - 2/2 - 2/ = 0, (d) s^ -f-s + j, + 3 = 0.
75. Knd the amplitude, period, frequency and epoch of the fol-
lowing S.H.M.
2/ = 7 sin 6i.
2/ = 6 sin 2irt.
y = a sin {id -f- e) .
76. Find cis= e. Show that
cos 5x = cos^ a; — 10 cos' x sin^ i + 5 cos x sin'' x.
77. Find graphically (on form MZ) the fifth roots of 2^ cis 35°.
78. Complete the following equations :
sin (a + b) = ?
tan 2x = t
cos (a ± 6) = ?
cot 2a; = ?
tan {a + B)='i
sin^ = ?
20
450 ELEMENTARY MATHEMATICAL ANALYSIS [§256
sin 2a; == ? cos - = ?
2
cos 2x = "> cot - = ?
2
79. Solve a' + 1 = 0.
80. y = — St' + 4i — 5 and x = 5t are the parametric equations
of a curve. Discuss the curve.
81. Show that [rfcos e +i sin S)] [(r'(cos B' + i sin B')] =
rr'lcos (e + B') +i sin (e + 6')].
82. Two S.H.M. have amplitude 6 and period two seconds. The
point executing the first motion is one-fourth of a second in advance
of the point executing the second motion. Write the equations of
motion.
83. Show that sin 5x = sin* a; — 10 sin' x cos' x -\- 5 sin x cos^ x.
CHAPTER XV
APPENDIX
A REVIEW OF SECONDARY SCHOOL ALGEBRA
300. Only the most important topics are included in this review
Prom five to ten recitations should be given to this work before begin-
ning regular work in Chapter I.
With the kind permission of Professor Hart, a number of the exer-
cises have been taken from the Second Course in Algebra, by Wells and
Hart.
, 301. Special Products. A few simple muItipUcations may be per-
formed mentally.
(1) The product of the sum and difference of any two numbers:
(a -I- 6)(a - 6) = o2 - 62
From this we have (3a; - 2y)i3x + 2y) = 9x^ - ^y^.
Exercises
Multiply mentally the following :
1. (3a; - J/) (3a; -|- y). 6. (29)(31), or (30 - 1)(30 + 1) =
900 - 1 = 899.
2. (2a; + 7)(2a; - 7). 7. (51) (49).
3. (5a; - y){5x + y). 8. (52)(48), or (50 + 2)(50 - 2).
4. Ixh/ - 3a){x'y + 3o). 9. (103) (97).
6. (o + 3b) (o - 36). 10. (25) (35).
(2) A few products of binomials are: \
(o + by = a' + 2ab + 6'.
(o - by = o' - 2o6-|- 6^
(a + 6)3 = a> + 3a'b + 3ab' + b\
(a - 6)» = a' - 3a'b + 3a¥ - bK
la + h)* = a* + 4o»6 + 6a'b' + 4o6» + 6*.
(a - b)* = a* - 4o»6 + 6a'b' - 4a6' + 6".
Thus (3 - o) 3 = 27 - 27o + 9a» + a\
and (a; -t- y^Y = x* + 4a;'v«+ Qx^y* + ^y^ + y\
451
452 ELEMENTAE/y MATHEMATICAL ANALYSIS [§301
Expand mentally the following:
1. (2o - x)K 4. {x - d)*.
2. (a; + 3yy. 5. (1 - a;)'.
3. (2x - 1)'. • 6. (2 + yY.
7. (52)2, or (50 + 2)', or 2500 + 200 + 4 = 2704.
8. (31)2, or (30 + ly. 9. (29)', or (30 - 1)^.
(3) The square of a polynomial is illustrated hy the following:
(a + b + c)2 = a" + 62 + c2 + 2ab + lac + 2bc.
(o + 6 + c + d)2 = a2 + 62 + c2 + d2 + 2o6 + 2oc + 2ad + 26c +
26d + 2cd.
(3 - a; + !/)2 = 9 + x2 + 2/2 - 6a;+ 62/ - 2xy.
Expand mentally the following :
1. (o + 6 + 2)2. ■ • 4. (2a - X + 3)2.
2. (a + 6 - 2)2. 5. (x2 - 22/2 + 4)2.
3. (a - 6 - c)2. 6. (x - 2o - 62/2)'.
(4) The product of two binomials having a common term:
(x + a){x + 6) = x2 + (o + b)x +,ab.
Thus (x + 5)(x - 11) = x2 + (5 - ll)x + 5( - 11),
= x2 - 6x - 55.
(x +7)(x + 2) = x2 +9x + 14.
(x - 5)(.x + 3) = x2 - 2x - 15.
(x2 - 22/) (x2 - 32/) = x" - 5x22/ + 6j/2.
Find mentally the value of each of the following :
1. (x + 2)(x + 3). 6. (3x + 22/)(3x - 7y).
2. (x - 2)(x + 3). 7. (x2 - 3)(x2 - 4).
3. (x - 2)(x - 3). 8. (3x1/ - z)(3x2/+ 7z).
4. (x + 2)(x - 3). 9. (x22/2 - 3)(x22/2 - 10).
5. (x2 + 52/) (x2 - 52/). 10. (x - 2y){2x - 2y).
(5) rfte product of two general binomials:
{ax + 6) (ex + d) = ocx2 + (6c + ad)x + bd.
Thus
(3o - 4b) (2a + 76) = (3a) (2a) + (- 8 + 21)ab + (- 4 b) (7b)
= 6a2 + 13ab - 28b2.
Find mentally the following products:
1. (5x - 22/)2. 4. (2m + 3)(m + 4).
2. (a + llb)(a + 36). 5. (2/2 + 4z)(2/2 + 4z).
3. (a - 2u)(a + 12»). 6. (3x2/ - 7)2.
§302] REVIEW OF SECONDARY SCHOOL ALGEBRA 453
7. (Sw^w - 4:)(.3uh; + 4). 29. (2 - 3s«)(5 + 2st).
8. {2x - 5)(a; + 4)i 30. (a^b + 6c) (0^6 - 13c).
9. (2r2 - 7)(3r2 + 5). 31. [Ip + 5)(lp - 4).
10. (p2 - Sq){,p^ + 7q). 32. {a' + 7)(o' - 11).
11. (a + l)(o - i). 33. (So + 5) (7a - 8).
12. iix + 5y){ix - By). 34. (1 + 8n)(l - 9n).
13. (u - |)(w - I). 35. (2a - 6") (2a + 3b*).
14. (2a; + 3)(Js + 1). 36. (12a; - i){9x - J).
16. (3x2 4. 46c) (3x' - 46c). 37. (20 - 16z)(3 + 2z).
16. iy -8)(.y + 5). 38. (r^ + 16s) (r^ - s).
17. (X - i){x - f). 39. (a - 6x2) (a 4. ^2^
18. (1 - 6s) (3 + 2s). 40. (4r + uv)(ir - 5uv).
19. (2< - '7w^){3t - 4u)2). 41. (6x2 _ 1)2,
20. (|u - i)(f« + J). 42. (1 + 23n)(5 - n).
21. (3r - 7<)(5r + 2t). 43. (x* - 2/*)(x« + y*).
22. (11x2 _ I)(i2x2 + 1). 44. (5a2 - 4b)(6a2 - 56).
23. (z2 - 6) (02 + 12). 46. (x2j/ + yH){xh/ - y^x).
24. (x + 32/2) (x - 22/2). 46. (fa + 10) (2a + 1).
26. (6m» - 6s2)(5m' + s2) 47. (9r + 2s) (3r - 4s).
26. (5x + |)(5x -i). ■ 48. (12x2 4. 5) (43,2 _ 3).
27. (3x + 7)(x - 5). 49. (a26* + 4x2)2.
28. (4o - 363)2. 60. (a^ - 5«)(a« + 6«).
61. (a + 6)(a - 6)(a2 + 62)(a* +6<).
302. Symbols of Aggregation. If a sign of aggregation is preceded
by the negative sign, change all signs within when the sign of aggre-
gation is removed. If the sign of aggregation is preceded by the posi-
tive sign, all signs within remain the same when the sign of aggre-
gation is removed.
5x2 _ [syi 4. {2x2 _ (2,2 4. 3^2) 4- 5j/2} _ ^2;
= 5X2 _ [■^yl 4. {2x2 - y2 - 3x2 4. 52^2} _ 3;2]
= 5X2 _ [3y2 4. {42,2 _ 3.2} - x2]
= 6x2 _ [3y2 _|_ 4j,2 _ -j2 - x2]
= 5X2 _ [7j,2 _ 2x21
= 5x2 _ 7y2 4. 2x2
= 7X2 _ -Jyl
Exercises
Simplify the following by removing the signs of aggregation :
1. ab - 46^ - (2a2 - 62) _ { _ 502 4. 206 - 862).
2. X - { 2/ -I- z - [x - ( - X - 2/) -f z]) + [z - (2x - 2/)].
454 ELEMENTARY MATHEMATICAL ANALYSIS [§303
3. o-{ -o — [-o-(-o- 1)]).
4. Syz - [2yz + (9z - 2yz)].
6. - { -1 -[-1 -(-1)]1.
6. 5x' - [Sy' + {2x^ - {y' + Zx'') + 5y^} - x'].
7. ab - [46^' - (2o» - b^) - [ - 5o' + 2ab - 3b']].
8. 33/' - I2y' + (9z - 2yx)].
303. Factoring. Since (a + b)' = a' ± 2ab + b', any expression
of the form of the right-hand side can be factored by inspection.
Thus,
x' - 6xy + 9y' = {x - 3y)'
and
4 + 4(o + 6) + (a + 6)2 = (2 + o + 6)'
Exercises 1
Factor the followlag by inspection:
1. Qx^ - ZQxy + 25yK
2. 4 + 16« + 16«».
3. a;*j/* + 10a;'j/2z2 + 25z*.
4. 9 + 6(x» + j/») + (I' + !/>)». .
6. a* + 4o26« + 4b<.
Since (o + 6) (o — 6) = a' — 6', any expression of the form of the
right-hand side can be factored by inspection. Thus,
4o2 - 9b' = (2a -|- 36) (2o - 36).
Exercises 2 j
Factor the following by inspection:
1. x'j/' - «'• 4. 25 - 3a;'.
2. (o -I- 5)' - c'. 6. 81 - 625x*.
3. c^ - {a-\- 6)'.
Since (a 4- 6)(o + c) = o' + (6 + c)a + 6c, any expression of the
form of the right-hand side can be factored by inspection. Thus,
a;' -5a; - 14 = (a; - 7)(a; +2)
Exercises 3
Factor the following by inspection:
1. a;' + 7s + 10. 4. 9i' - 18s - 27.
2. a' + 4aj/ - 21?/'. 5. 25 + 30o - 27a».
3. 4a;' - 18iy + 18i/'.
§303] REVIEW OF SECONDARY SCHOOL ALGEBRA 455
Since (a + &)(«' — ah + 6*) = a' + b', any expression of the form
of the right-hand side can be factored by inspection. Thus,
27 + 125a;» = (3 + 5x)(,Q - ISa; + 25a;2).
Exercises 4
Factor the following by inspection:
1. x'y' + 1. 4. 125 + x'yK
2. x' + y". 5. x' + 8yK
3. 8 + 27a;'. i
Since (o — 6)(a^ + ab + 6') = o' — 6', any expression of the form
of the right-hand side can be factored by inspection. Thus,
27 - 125a:' = (3 - 5x)(9 + 15a; + 25ai2).
Exercises 5
Factor the following by inspection:
1. x>y' - 1. 4. 125 - x^yK
2. x^ — y^, or (a;* + y^)ix' — y^). 5, z^ — 8yK
3. 8 - 9a;'. 6. 27 - 8a';
The following may be factored by grouping the terms. Thus,
a'm + o're — m — n = a'(m + n) — (m + ra)
= (o' — l)(m + n)
= (a - l)(a2 +a + l)(m + n).
Exercises 6
Factor the following:
1. ax — ay + bx — by. 4. x^ — xy* — x^y + y^.
2. a;' + 3a2 + 3a; - 1. 5. a;* - x^y - xy^ + y\
3. ax^ - 2axy + ay' + bx' - 2bxy + by',
A trinomial of the form px' + gx + ?■, if the product of two bino-
mials, may be factored as outlined below.
In the product
ax + b
ex + d
OCX' + {be + ad)x + bd
456 ELEMENTARY MATHEMATICAL ANALYSIS [§303
the terms acx'' and hd are called end prodiicts and bcx and adx are called
cross proditcts. This most important case of factoring is best learned
from the consideration of actual examples.
Factor 21x'i + 5a; - 4.
Prom the term 21a;*, consider as possible first terms 7s and 3a;, thus
(7a; )(3a; ). For factors of (— 4), try 2 and 2, with unhke signs,
and signs so arranged that the cross product with larger absolute
value shall be positive; thus (7a; — 2)(3i + 2). This gives middle
term Sx; incorrect. For (—4) try 4 and 1, with signs selected as be-
fore; thus, (7x — l)(3a; + 4). Middle term 25a;; incorrect. Try
(7a; + 4) (3a; — 1). Middle term 5x; correct.
Factor 2ix' - 17xy + ZyK
Try (6a; — 32/)(4x — y). Incorrect, since first () contains factors
and given expression does not. Try (fix — y){4:x — 3y). Middle
term - 22; incorrect. Try (8a; - 3y)(3x - y). Middle term - 17;
correct.
Exercises
1 7
Factor the following:
1. 6x' - 7a; + 2.
8.
35u2 + UV - 6t)2.
2. 3x2 -i- 8x + 4.
9.
9*2 - 14* - 8.
3. 6x2 - a; - 2.
10.
121^^ - 35x2/ - 32/2.
4. 9a2 + 15o + 4.
11.
6 - i - 15*2.
5. 66" - 76 -10.
12.
5 + 9s - 18s2.
6. 14x2 + 13^ _ i2y\
13.
24m2 - 17mn + 3n2.
7. 8z2 - 2yz- 2lyK
14.
28y" - yz - 2zK
An expression of the from o* + 0252 -\- b* may be put in the form of
the difference of two squares by adding and subtracting a^b'. Thus,
a* + a'b' + b< = a* + 2a2b2 + (,2 _ a^b^
= (a2 + 62)2 _ a'b"
= (a2 + ab + 62) (a2 - ab + b').
Exercises 8
Factor the following:
1. X* + x'y' + yK 5. 16x* + 36x'y^ + 81yK
2. X* + 4x2 4. 16. 6. a* + a*V + b*.
3. 2/* + iy'z^ + I62 . 7. aV + a^x'y' + y\
4. 16 + 4«2 + u\ 8, 625x« + 100x2z< + 162». .
§304] REVIEW OF SECONDARY SCHOOL ALGEBRA 457
304. To factor a polynomial completely, first remove any monomial
factor present; then factor the resulting expression by any of the type
forms which apply, until prime factors have been obtained throughout.
'Thus,
(a) 5a« - 5&« = 5{a^ - 6«) = b{cfi - V){a^ + 6=)
= 5{a - b)(a2 + a6 + V){a + h){a^ - ah -^ h")
(b) 42aa;2 + lOox - 8a = 2a{2\x^ + 5x - 4)
= 2a(Jx +4) (3a; - 1)
(c) 2Cmnu^ - IWmnu + X2imn = 5mn{^' - 20m + 25)
= bmnhu - 5Y.
Exercises
Factor the following expressions: ,
1. xV"° - A"*- 22. a;2 + Qx - 27.
2. 9x» - 43/6. 23. c' -64«3.
3. ,25a;« - 1. 24. Sx' - 1.
4. 81 - ^K 26. 1 - 13< - 68«2.
6. 1 - 6ia''b*c\ 26. a;< - Cx^b - SSb".
6. a;' — y^. 27. au" — 4aM!; — i5av^.
7. 225 - aS. 28. 28a2 - a - 2.
8. 121x2 - 1442/2. 29. Ss^ - 17si + 24{2.
9. 49ot« - SQx'y^zK 30. 15r= - r - 6.
10. 169 -:^ a'lx^. 31. iy^ - 3y - 7.
11. 4x2 _ 20x + 25. 32. 641*6 _ 27x3.
12. 9o2 + 6ob + b2. 33. 6ar - 3as + 4a«.
13. a'b^ - nabc - QOcK 34. a^ +2a - 35.
14. r* - llr' + 30. 35. 9x2 ^ i2xy - 32^".
16. 16b2 + 30b + 9. 36. o" + lOab + 25b2.
16. Slu" + 180ua + lOOs;' 37. 625x22/2 - ^.
17. 36a2 - l32o + 121. 38. 3cdy' - 9cdy - 30cd.
18. x'y* - Axy^ + 4. 39. 4ox2 - 25ay*.
19. o2b2 - 2ab - 35. 40. 3y^ +24.
,20. u' +~u3 _ 110. 41. 4x2 _ 27x + 45.
21. a*b2 - 14o2b + 49. 42. 6x2 + 7^ _ 3,
458 ELEMENTARY MATHEMATICAL ANALYSIS [§305
43. -jV' - 1. 58. 2am« - .50a.
44. 10x>y - 5x^y^ - 5xy\ 69. 72 + 7a; - 49a;».
45. »i»n» + 7mn - 30. 60. 31a;' + 23xy - 8yK
46. x^ - Zxy - 70yK 61. 24o» + 26a - 5.
47. mx" + 7mx - 44r«. 62. 1 - 3xy - IQSx'y^
48. x' - 3a;» - 108x. 63. x^ - Umx + AOm^.
49. x> - yK 64. 26 + 10a5 - 28o%.
60. a;* - hx-^ - 'iAy\ 65. c» + 27(f».
61. 8n« + 18n - 6. 66. Zx^y - 27xy\
62. 3i* - 12. 67. -^^^ - 4^^*-
63. Stw" - 42ot« + 49<». 68. 49ji<2/ - 196nV-
64. lOa;' - 39a; + 14. 69. a;« - 16a; + 48.
65. 12x« + 11a; + 2. 70. a;' + 23a; - 50.
66. 363;" + 12i - 35. 71. a<w« + 31a!^2 + 30.
67. x' - SyK 72. 9a;' + Z.7xy + iy\
306. General Distributive Law in Multiplication. From the mean-
ing of a product, we may write
(a + 6+c+. . .){x + y + z + . . .)=ox + 6a; + ca;-|-. . .
+ ay + hy + cy +. . .
+ az + bz + cz +. . .,
etc.
Stating this in words : The product of one polynomial by another is the
sum of all the terms found by multiplying each term of one polynomial
by each term of the other polynomial.
To multiply several polynomials together, we continue the above
process. In words we may state the generalized distributive law of
the product of any number of polynomials as follows:
The product of k polynomials is the aggregate of all of the possible
partial products which can be made by multiplying together k terms, of
which one and only one must be taken from each polynomial.
Thus,
{a + b+c-V. . .){x+y + z+. . .)(«+« + 10 +. . .)
= axu + axv + . . . + ayu + ayv + . . . + azu + azv + . . .
+ bxu + bxv + . . . + byu -H byv -|- . . . + bzu + bzv + . . .
+ cxu + cxv + . . .
4- . . .,etc.
If the number of terms in the different polynomials be n, r, s, t. . .
respectively, the total number of terms in the product will be nrst , , .
The student may prove this.
§306] REVIEW OF SECONDARY SCHOOL ALGEBRA 459
306. The Fundamental Theorem in the Factoring of z" ± a". The
expression (s" — a") is always divisible by (x — a), when n is a posi-
Write a"— a» = s" — ox"~i + oa;""^ — a"
= a;"~'(a; — a) + oCa;"""- — o"~^)
Nowi/(a;'~' — o'~i) is divisible by {x — o), then'plainly s'~'(a; — a)
+ a(a;'~' — a'~i) is also divisible by {x — a). But this last expression
equals (x* — o*), as we have shown. Therefore, if (x — a) exactly
divides (a;*~i — o*~'), it will also exactly divide (a* — o*).
That is, if the law is true for any positive integral value of k, it
is true for k one greater. But by actual division the law is true when
k is 3, (x' — o' = (x' + ax + a'){x~a) therefore it is true when
k is 4. Being true when k is 4, it is true when k is 5, and so on up
to fc = n, any positive integer.
We see that (x — o) is one factor of (x" — o"). The other factor of
(x"— o») is found by actually dividing (x" — a") by (x —a). Thus
(x» - a") = (x - o)(x"-i + ax"-' + a'x"-^ + . . . + o»-=x + o"-')
The student may show that (x + a) divides x" + o" if ra be odd, and
divides x" — a" if n be even.
Exercises
Factor the following:
1. x' + yK 7. m* - 243.
2. x5+32. 8. 32o» +2436^
3. x« - 81. 9. 64 - x«.
4. x» + 1. 10. x'y' -z».
6. X* - 162/*. 11- a:' - 2/".
6. x'j/s + 1. 12. 27x5 - 8y\
Miscellaneous Exercises in Factoring
Factor the following:
1. {a+by - c\ 10. (1 +«!=)«- iuK
2. (to - n)2 - x\ 11. 9(to - n)" - 12(m - n) + 4.
3. (x - 2/)2 - z2. 12. (xii - 4)2 - (x + 2)2.
, 4. x2 - (v - z)2. 13. (x2 + 3x)2 + 4(x2 + 2x) + 4.
6. (7x - 2yY - y'. 14. (Ox^ + 4)2 - 144x2.
6. (a + 6)2 + 23(a + b) + 60. 15. (x + y^ + 7(x + 2/) - 144.
7. (x + y)' + 2(x +y) - 63. 16. (a2 + o + 9)' - 9.
8. (x - yy - (x + 2/)2. 17. (x + 2/)» - z^
9. (x' - 22/)2 + 2(x2 - 22/) + 1. 18. (x + y)' + z».
460 ELEMENTARY MATHEMATICAL ANALYSIS [§307
19. x^ - {y + z)K 33. 9a* - ix^ +.j/»- 6x'y - 20x« -
20. x^ - \y - z)'. IhzK
21. aj= + {y - 2)3. 34. 4i/* - 322/2 + 1.
22. (to + «)s + 8«'. 36. 94" - 31««x« + 25a;*.
23. (re + y)' + (a; - y)'. 36. 25o* + 340^62 + 496*.
24. 27a» - (o - b)». 37. 2a(a; + i/) - 3(a; + y).
26. o' - 2a6 + 6* - c^ 38. a(s - j/) - 6(a; - y).
26. a;« + 2xy + 2/^ - z^. 39. ab + on + 6to + mn.
27. a^ - a;2 - 2a;2/ - j/^. 40. 2 + 3a; - Sa;^ - 12a;3.
28. a;2 - 2/2 - z2 + 2j/z. 41. 56 - 32a + 21a2 - 120^.
29. b" - 4 + 2a6 + 02. 42. 4a' + o^b^ - 4b= - 16ab.
30. 2mn - m^ + 1 - m^. 43. si - sr - r" + r«.
31. 9a2 - 24ab + 16b« - '^e. 44. a;^ + a;2 + a; + 1.
32. 4a2 - 6b - 9 — b^.
307. Fractions. Multiplying or dividing both numerator and
denominator of a fraction by the same number, excepting zero, does
not change the value of the fraction. To reduce a fraction to its low-
est terms factor both numerator and denominator and then divide out
the common factors if there are any. Thus,
ai*+ aa;2j/' + a2/* _ a{x'^ + xy + y')(3;' — xy + j/') _ x' — xy + y'
aV — a'y^ a'(x — y)(x' + xy + y') " a(x — y)
Exercises
Reduce the following to lower terms:
ax + ay — X — y
1.
ax^
- ay^
aV
-a'y^
2.
a;» +
X -6
x^
- 4 ■
3.
X* -
y
4.
6.
6.
x^ + y^
27 -X'
12 - 7x +a;2-
as - b*
a" - V'
Miscellaneous Exercises in Fractions
1.
SimpUf y the following :
x^ - 36 7n2
4n2 n^ +n - 42
„ x' - 2x - 35 4x2 _ Qx
' 2x8 _33;2 ■ 7(^ _7)-
„ (5a +2) (a -2)
• 2a2 + a - 10 ■
§308] REVIEW OF SECONDARY SCHOOL ALGEBRA 461
4o2 + 8a + 3 6a2 - 9o
i.
6.
2o2 - 5a + 3 4a2 - 1 ■
16a; - 4 _ 20a; + 5 a;^ + 2a: + 1
5a; - 5 ' 6a: + 6 " I6x^ - 1 '
a:» + 8y^ _ X -2y _ a;" + 23;;/ + 4^^
a;' - 8^' ' a; 4- 22/ ' x^! - 2a;2/ + iy"'
2n' -n - 3 n^ + 4»i + 4 ra^ -
10.
n* - 8n2 + 16 n^ + n 2n' - 3n
x^ — xy — 2y^ ^ x — 2y '
x' — 9xy^ ' X — 3y
2a;g -xy - 3y' ___ 3x^ + xy - 2y'
9x' - 25y' ' 9x' - 30a;^ + 25^8'
2a' - Sab - 36= . r2a' - 7ab - 46= . a' - 4^ab + 4an
■ L a' -
a2 - o6 - 262 ■ la' - 3ab - 46' ' o^ - ab - 66^
3 \
/^+2 _x_\ /^^
V X ^ X - 3/ \a: - 2 x + 3/
1
X
308. Simple Equations. Adding the same number to both mem-
bers of an equation does not change the equation. It follows that a
term may be transposed from one member of an equation to the
other member provided its sign is changed. Thus from
3x - 2 = 3 + 2x.
3x - 2x = 3 + 2, or X = 5.
Exercises
Solve the following equations for x:
1 ^ ~ ^ 4. 2x - 1
X -3
, 2x - 1
"•" 4x - 3
2x + 1
6x +7
3
3
X + 2
X -2
2x + 3
2x + ■.
o a; -p ^
„ X - 2 2x + 3
^- ~3 r~ - "■
4. (x + 4)(x - 2) = (x + 3)(3x + 4) - (2x + l)(x - 6).
6. (,v' -2x + ly = (x - iy{x - 3)".
462 ELEMENTARY MATHEMATICAL ANALYSIS [§309
309. Quadratic equations are usually solved (a) by factoring, (6)
by completing the square, or (c) by use of a formula.
(a) To solve by factoring, transpose all terms to the left member of
the equation and completely factor. The solution of the equation is
then deduced from the fact that if the value of a product is zero, then
one of the factors must equal zero. Thus
(1) Solve the equation
a;2 + 54 = 15a;
Transposing x^ — 15a; + 54 = 0
Factoring (a; - 9) (a: - 6) =0
a!-9=0ifa;=9
X - 6 = Qiix = 6
Hence the roots of the equation are 9 and 6
Check: Does (9)' + 54 = 15 X 9?
Does (6)« + 54 = 15 X 6?
(2) Solve the equation
12a;2 + x = &
Transposing 12a;'' + a — 6 =0
Factoring (3a; - 2) (4a; + 3 = 0
3a;-2=0ifa; = |
4a; +'3 = 0 if a; = -f
Hence the roots of the equation are f and — f.
Check: Does 12(f)2 + f = 6?
Doesl2(-f)»-f=6?
(b) To solve by completing the square, use the properties of
(a; ± a)* = a;' ± 2oa; + o*, as follows:
REVIEW OF SECONDARY SCHOOL ALGEBRA 463
(3) Solve x^ - V2.X = 13.
Add the square of 1/2 of 12 to each side
a;2 - 12a; + 36 = 49
Take the square root of each member
a -6 = ±7
Hence
a; = 6 + 7 = 13
a; = 6 -7 = -1
Check: Does (13) ^ - 12 X 13 = 137
Does (-1)2 - 12 X (-1) = 137
Since in general (a; — a){x — V) = a;^ — (a + b)x + o6, we can check
thus:
Does 13 + (- 1) = - (- 12)7
Doesl3(-l) = -137
(4) Solve x^ - 20a; + 97 = 0.
Transpose 97 and add the square of 1/2 of 20 to each side:
a;' - 20a; + 100 = - 97 + 100 = 3
Take the square root of each number:
a; - 10 = + -\/3
Hence
si = 10 + -\/3[
a;2 = 10 - ^3
Check: Does xi + Xj = — (— 20)7
Does xiXz =97?
(c) To solve by use of a formula, first solve
0x2 + 6x + c = 0 (1)
The roots are
- h±y/V - 4ac
2a
(2)
For a particular example, substitute the appropriate values of a, 6,
and c. Thus:
(5) Solve 2x2 - 3x - 5 = q.
Comparing the equation term by term with (1) we have
o = 2, 6= -3, c= -5
464 ELEMENTARY MATHEMATICAL ANALYSIS [§309
Substitute these values in the formula (2)
^ _ -(-3) + V(-3)'-4(2)(-5)
2(2)
3 + 7
Therefore
xi — 5/2, X2 = — 1
Check: Does Xi + x^ = — b/a = 3/2?
Does X1X2 = c/a = — 5/2?
Exercises
Solve the following quadratics in any manner:
1. s= + 5x + 6 = 0. 29. 3i2 _ I2ax = 63a'.
2. a;2 + 4a; = 96. 30. ix' - I2ax = 16a^
3. x' = 110 + X. 31. x^ - X =6.
4. x' + 5x = 0. 32. x' +7x = - 12.
6. 6x2 + 7a; + 2 = q. 33. x' - 5x = 14.
6. 8x2 - lOx + 3 = 0. 34. x^ + x = 12.
7. x' + mx - 2m,' = 0. 36. x" - x = 12.
8. 3«2 - « - 4 = 0. 36. x' = Qx - 5.
9. 107-2 + 7r = 12. 37. x' = - 4x + 21.
10. x' + 2ax = b. 38. x^ = - 4x + 5.
11. x2 + 4x = 5. 39. x2 + 5x + 6 = 0.
12. x2 + 6x = 16. 40. x' + llx = - 30.
13. 2x2 - 20x = 48. 41. x2 - 7x + 12 = 0.
14. x2 + 3x =18. 42. x2 - 13x = 30.
15. x2 + Sx = 36. 43. 3x2 + 4^ = 7.
16. 3x2 4. 6x = 9. 44. 3x2 + 61 = 24.
17. 4x2 _ 4a; = 8. 45. 4-^2 - 5x = 26.
18. x2 - 7x = - 6. 46. 5x2 _ 7^; = 24.
19. x2 - ax = 6a2. 47. 2x2 - 35 = 3x.
20. x2 - 2ax = 3a2. 48. 3x2 _ 50 = 5x.
21. x2 - X = 2. 49. 3x2 _ 24 = 6x.
22. x2 + X = a2 + o. 50. 2x2 - Sx = 104.
23. x2 - lOx = - 9. 51. 2x2 ^ iqx = 300.
24. 2x2 _ 15a; = 50. 52. 3x2 _ iqx = 200.
26. x2 + 8x = -15. 53. 4x2 - 7x + ^ = q.
26. 3x2 + 12x = 36. 54. |x2 _ f x = - ^^.
27. 2x2 + lox = 100. 55. 9x2 + 6x - 43 = 0.
28. »2 _ 5a; = _ 4, 66. 18x2 - 3x - 66 = 0.
§309] REVIEW OF SECONDARY SCHOOL ALGEBRA 465
67.
|x2 - 3i + il = 0.
59. 2x2 _ 22x = - 60.
68.
X* 3a;
4-1+2=0.
60. 3x» + 7x - 370 = 0.
61. 5x' -ix-^Ts = 0.
* - x^
x
1
62.3
-2+i- = o-
63.
x' + 2x + I = Qx + 6.
69. s2 = 5s + 6.
64.
s^ - 49 ^ 10(a; - 7).
70. r' + 3r = 4. --^
66.
2x' + 60a; = - 400.
71. 2s2 + 4as - c = 0.
66.
qs + 7a + 7 = 0.
72. x" + 6ox - 5 = 0.
67.
z' = 3z + 2.
73. x2 - lOax = - 9oa.
68.
r = r2 - 3.
74. ca;2 + 2ix + e = 0.
75. 2x-
' + 6x
- n = 0.
76.
y' + h = l
83. 4x2 - 3x = 3
77.
x^ = 6 + 4a;.
84. W + 4< = 6.
78.
M« - fu - 1 = 0.
85. 5(x2 - 25) = X - 5.
79.
t'+it= I
86. 9i42 + 18m + 8 = 0.
80.
r» - f = ir.
87. x* + px + 3 = 0.
81.
s^-is= ^.
88. X* - 8x2 + 15 = 0.
82.
3r' -2r = 40.
89. M* - 29m2 + 100 = 0.
2 8
5 8
90.
X' ~ 3x'
93.5_, + 8f,-3.
91.
^y + i = ly
n-3 n+i_
n— 2 n 2
92.
* + 4a; ix'
95.3x + ^^ + ^^=2 +
X
- 24
24
96.-
X —
2 + 1=0.
97. x«
- 35x=
+ 216 = 0.
98.
.■ + f —
99. (a
, 1\2 16/ , 1\ , ^
'+x) - 3-(^+xj +7 =
= a -\
100.
u
a
101
2a; - 7 lOx - 3
33
i
x2 - 4 5x(x + 2)'
Hint: Clear of fractions by multiplying both members of the equa-
tion by 5x(x2 — 4), the lowest common denominator.
102. — ? I '^~ = 0
2x + 1 3x + 2 ^ 6x2 + 7x + 2 "■
103.
12x - 5 3x + 4 4x - 5
21 3(3x + 1) 7
104. ^' +3 , 1 _ 2x - 1
2(x' - 8) 6(x - 2) 3(x2 + 2x + 7)'
3Q
466 ELEMENTARY MATHEMATICAL ANALYSIS [§309
106. _£_4.a:-l_a;»+a;-l
106.
X ~ 1 X x' — X
X x__ _ a:' + 23: — 2
s+2 a; + 3~a;' + 5a;+6*
107. ~ 1 Z?_ L ^^ = 0
a; - 2 ^ 24(a! + 2) ^4 - a;^
108. (. + -!)= -^(a;+g + 7 = i
Hint: Let a; + - = j/. Then
2/' - Y!/ + 7 = *■
Solve this equation for y. Place a; + - equal to each value found
for y and solve the resulting equations for x. There are in all four
roots of the given equation.
109. a;« - 35a;« + 216 = 0. ,,, _i_ 1 o i i
Hint: Let a;' = v. 111. a: + - = 2 + 2.
110. .^+ ^0^29. ii2.?i__24 +1=0.
a;" a; a; — 2
113. (a;2 + a;)2 = 12 + ^{yfl + a;),
a; + 1 12(a - 1)
114. 1 +
116.
a;-l a; + l
Ca;-l)(a;-2) _(a; + l)Ca; + 2)
X -Z » + 3
116. Solve oV _ 2fa2/ + bV = 1 for J/i considering a, &, ^, and a; as
known numbers.
307. The Definitions of Exponents.
(1) n a positive integer: o" = aaa . . . to n factors.
(2) n and r positive integers: a^' = '^ a and a"/' = (Vo)"
= v^.
(3) o» = 1.
(4) n any number, positive or negative, integral or fractional:
a-» = l/o».
308. The Laws of Exponents. For n and r any numbers, positive
or negative, integral or fractional :
(1) a"o' = o"*', or law for multipUoation and division.
REVIEW OF SECONDARY SCHOOL ALGEBRA 467
(2) (o")' = a", or law for involution.
(3) o'S" = (aft)", or distributive law of exponents.
Note: The student must distinguish between — a" and (— a)".
Thus - 8!^ = - 2, and (- 8)!^ 2, but (- ZY = 9 and - 3' =
-9.
Exercises 1
Use the definitions of exponents (1), (2), (3), (4) §307, and the laws
of exponents (1), (2), (3), §308, and find the results of the indicated
operations in the following exercises.
1. x"x".
9. x" -T- i«.
17. (a')3.
2. a'^'a"'.
10. a" -f- a".
18. (o*)».
3. a;i'»+ii».
11. o'» ^ a».
19. (-o62)».
4. 626»+s.
12. e»+' -7- e».
20. (o»2/«)s.
6. W+'m""'.
13. IC'+s -=- 10'.
21. (ft") 2.
6. o"-»o»+».
14. n'+» -f- n'+3.
22. (-o»6'>'.
7. s»-'+V.
15. u""'^ -^ u""'.
23. (a«6»)'.
8. m^'^mr''.
16. aj'-f+i ^- a;'.
24. (r'»s»)''.
- ©'■
Exercises 2
27. (-
Write each of the following sixteen expressions, using fractional
exponents in place of radical signs:
1. v^.
5. v^
9. v^iT
13. V'o-5.
2. V^-
6. (v^=.
10. (>^3.
14. (-C^a - 6)'.
3. Vc^.
1.-^.
11. </x'.
16. -s/a? - 6''.
4. v^
8. {<fS)K
12. (^^^
16. y/ia+hy.
Find the numerical value of each of the following sixteen
expressions:
17. 4*.
21. 625*
25. 81^.
29. 256*.
18. 27*.
22. 64*.
26. 125^.
30. 64^.
19. 9*.
23. 216*.
27. 32^.
31. 512^
20. lei.
24. 16^.
28. 81^.
32. 128*.
Write each of the following expressions in two ways, using radical
signs instead of fractional exponents:
468 ELEMENTARY MATHEMATICAL ANALYSIS [§309
S3, o*.
37.
n^. 41. r*.
46. ol.
34. 1*.
38.
b*. 42. xX
46. 6"^.
36. TO*.
39.
^. 43. r-
n+1
47. a; - .
36. s*.
40.
&i 44. ol
Exercises 3
48. 0 '•
Perform the indicated operations in each of
the following examples
by means of the laws of exponents.
1. a? X a*.
a*
X a* = a«+* = a^^A =
= o^.
2. x^ X a;i
4. a;i X s*.
1 4
6. X :•> X a 8n
3. x^ X K*.
5. o* X o^.
8. (^ H- a^.
S r
7. o» Xai:.
a^
^ at = a^t = a4*-M .
= aA.
9. h^ 4- hi.
11. Sa^ftt H- 4o26^.
13. 6at -=- 3a*.
10. nfi -H vi
A.
12. 9a* H- a*.
15. (a^)^.
14. a6T -=- o 6"C.
(a^) ^ = o^ = oT^.
»
16. (o*)A.
18. (a*)^.
20. [(x»)f]f.
17. (h^^.
19. (a*)*.
22. (ata;i-2/t)i
21. (si^)r.
(a^x^y^
J)*=(at)W)*(2/¥ =
; a*x*j/i''.
23. (o^fe*)*.
26. (36a*a;22/')*.
27. (32x%4)*.
24. (,adi)i.
26. (a^a;^2/*)».
28. (^a'6'c)*.
- (i)*-
/aiy_(af)i a*
VftV (6*)^ 6i
-(?;
-(i)'
-(^)'
"■©•
A'\*
-m
36. (a* + at -f l)(o* -t- a - cji).
§309] REVIEW OF SECONDARY SCHOOL ALGEBRA 469
We arrange the work thus:
J + J + 1
a'' + a — gi
ai +J + a
_ of _ o _ ai
a' + 2ai + a* - a^
37. (x + 22/4 + 32/*).(i - 2yi + 32/*).
38. (X* + yh(.x^ - yh-
39. (,J - Sah^ + 4:ah - ah^-)(ai - 2o*6*).
3 i_ 1 ^ ^
40. (a» - 20"+ 3o")(2a» - a").
Exercises 4
Find the numerical value of each of the following;
1. 2-1. 4. 10-5. 7. 2-\ 10. 1024-*.
2. 4-2. 5. l-» 8. 16-". 11. 512-4.
3. (-2)-». 6. 2-2. 9. 81-4. 12. 625-*.
1 5 5-2 16-i
13. i- 16. (:r^,- 1,7. —^- 19. ^3r-
9 . 1-8 32-4 7-1
14. J-. 16. I^i- 18. ^^iir- 20. — ^•
3-2 8 1 21 49-1
Write each of the following expressions without using negative
exponents:
21. x-K 25. 5a-'. 29. (a; + y)-\ 33. 2o«a;-^-4.
22. x'y-K 26.30-^6-4. 30. (- x)-^ 34. (- a^)-^
„o 1 „„ 2a-2 „, X* „^ a-ibi
23. ^Ti- 27. „,. _,- 31. -^- 35. n"^-
„. ■mr' „„ 0^6-* „„ 3(rl6-i „„ 3a26-2c-''
24- -^- 28. _■ _,■ 32. ^ — 5 — 36. g^,.,,,^,.-
a; " a-3^-5 so-fj, 5o ^b =c *
Write each of the following expressions in one line:
37 (1. 39 -^. 41 ^^- 43 ''"y" -
38 1. 40 ^^-^. 42 '^E^yll. 44 J??i
38- a' f • 3a-2r» *2- u>z-^ **" ^=F^
16(a+ b)-'c4. ^ , 1 , 1, A-.
"• (a - 6)-lc-« x3 + x=i + X + a-i
470 ELEMENTARY MATHEMATICAL ANALYSIS [§309
Exercises 6
Perform the indicated operations in each of the following by means
of the laws of exponents.
1. o« X o"". 4. 8a-* X 3a'. 7. m"* X m"*.
2. r" X r-i». 6. m-i X tt*. 8. Sax-' X kbx\
3. c-> -r c-K 6. a;= -=- a;-". 9. o-»6-» -r- ab"'.
10, (- 7o-»6-»)(-4oi'b-')(a-%''a;-').
11. (2o*6-*)(a-*6*- |o*b* + ah-^).
12. 7a-»b-V-'-i- 80-26-%-
13. S6a;5J/-'^4 -^ Tai-iy-'a-*,
14. 18o-i&lc-5 -=- 6a*bV«.
16. Cai'j^-^ai -r 2a;-^3/iz-i.
16. (a-')!.
17. (o-s)-".
18 (a«)-«.
19. (7i*)-».
20. (r-iyi.
21. (c-')^
22. (obc)-*.
33. (bj.
42. (a2a;-i+3a=x-i')(4o-i - 5a;-i + 6ax-^)
4a-i - 5a;-i + &ax-'
aH-^ + 3a'a;-'
4ox-i - 5a"a;-2 + 6o»a;-»
12a'a;-' - 15a»a;-° + 18a«z-«
4aa;-i + 7a2a:-2 - 9a'a;-= + 18a*a;-*
43. (2a;-* - 3a; + 4a;*) (a;-? - 2a;-* + 3a;-*).
44. (a;-* - 2a;-*^ + y^){x-^ - y^).
46. (3a;* - |x* + 4) X 2a;-*.
46. (a;-^ + x'h + l)(a;"* - 1).
47. (a;-*+ 3/-=) (a;-* - y-^).
48. {x^y + yh{x^ - y-*).
§309] REVIEW OF SECONDARY SCHOOL ALGEBRA 471
49. (2o*- 3axi){3a-i .+ 2a;-*) (4a*a;* + 9o-M).
60. (x-^ - x-iyi + x-iy - y^) -■ (a;-* - j/*).
x~^ — y^)x-^ — x-^y' + x-'y — y'(3r^ + y
x-^ — x-hji
x'^y — y'
x-^y —y'
Bl. (x-' + 2x-°- - Sx-i) -i- (x-\+ 3s-i).
309. Reduction of Surds or Radicals.
1. // any factor of the number under thi radical sign is an exact
■power of the indicated root, the root of that factor may he extracted and
written as the coefficient of the surd, while the other factors are left
under the radical sign.
(1) Thus,
VS = V4 X 2
= VW2
= 2v^
(2) Also,
</81 = V27 X 3
= V27i/3
= 3i/3
(3) Also,
Vl&ax* = i/Sx' X 2ax
= VSx'V2ax
= 2x^2ax
2. The expression under the radical sign of any surd can always he
made integral.
(1) Thus
(2) Also
30
(2
3
-'4x\'
= J— X 18
\27
= iVl8
3
(7.
i
*/7 2
= Vi X 2 =
^i^x"
=
\VT.
472 ELEMENTARY MATHEMATICAL ANALYSIS [§309
3. W& may change the index of some surds in the following manner:
(1) Thus, a/I = vVi
= V2
1,2) Also, VlOOO = -s/^^/W^Q
= Vio
(3) Also, V2563^ = v/VilPas
\
= v^lGca'
A surd is in its simplest form when (1) no factor of the expression
under the radical sign is a perfect power of the required root, (2) the
expression under the radical sign is integral, (3) the index of the surd is
the lowest possible.
Methods of making the different reductions required by this defini-
tion have already been explained. We give a few examples.
(1) Simplify ^^^^
(2) Simplify^—.
a= 'a 1 3,_-- -
*/400 /20 1 , 2 ^__
(3) Simplifying-^.
5 « /512 5 /"S h .„ -
2 \ 125 2 '
f4) Simplify 3\/2 + 2^- + Vs.
3^2 + 2-/- + VS = 3V2 + \/2 + 2^2
= 6V2.
§318] REVIEW OF SECONDARY SCHOOL ALGEBRA 473
In any piece of work it is usually expected that all the surds will
finally be left in their simplest form.
Exercises
Reduce each of the following surds to its simplest form :
5. ^ — 7. ^ .
5 \4 \81 \b'x^
\3 \27 \12 \ x'
9. Simplify V^ + AV^ + 6VS-
10. Simplify 1 + Vs + V2 - V27 - Vl2 + Vl5.
11. Sunplify ■i/21 + 7v^ X "^21 - 7V5.
12. Find the value of a;^ - 6a; + 7 if a; = 3 - \/3.
13. Find the value when x = \/3 of the expression
2a;- 1 _ 2a; + 1
(a;- 1)2 (x + 1)2'
14. Find the value of
(35VT0 + 77\/2 + 63-v/3)(vT^ + V2 + \/3).
Solve and check each of the following equations:
15. vT+4 = 4. 22. Vx - Vx -5 = \/5.
16. ■v/2a; + 6 = 4. 23. Va; - 7 = V a; - 14 + 1.
17. VlOa; + 16 = 5^ 24. Vx - 7 = -y/x + 1 - 2.
18. V2x + 7 = VSx - 2. 26. x = 7 - Vx" - 7.
19. 14 + -i^4x - 40 =^10. 26. Vx + 20 - Vx - 1 -3=0.
20. Vl6x+9 = 4V'4x - 3. 27. -y/x + 3 + ■\/3x - 2 = 7. _
21. V^- + X = f + Vx. 28. V2x+ 1+ Vx - 3 = 2Vx.
20x , . 18
29- -7?^?=^ - VlOx - 9 = -jrz : + 9.
VlOx - 9 VlOx - 9
30 ^^^ _ Vi + 1
31.
Vx — 1 X — 3
Vx + Vx — 3 _ 3_
Vx — Vx — 3 X — :
318. Rationalizing the Benominator of a Fraction.
Illustration: Rationalize the denominator of
V3 + 2 (V3 + 2)(V3+ V2) 3 + 2V3+V6 + 2V3
V3-V2 (V3 - V2)(V3 + V2)
474 ELEMENTARY MATHEMATICAL ANALYSIS [§318
ExerciBes
Rationalize the denominator of each of the following :
1 6 g Vl^^^ + 1
' 3 + -v/s" ' Vx -2+2'
2 VS - V2 ^ Vo - 6 + Va
' VB + -\/2 " Va — b — Va
g 2V2 + 3 g Vl + g - Vl - g
■ 3\/2 + 2 ' Vl +a + Vr^^
^ 5>/2^+6 g 1
3V2 - 6 ■ a;2 - Vl + x'
Vi + Va ' 1 + Vi - x^
4 76 ELEMENTARY MATHEMATICAL ANALYSIS
LOGARITHUB
IO|il3l3l4lSl6l7l8l9lia
3 U
5 617 8 9 1
10
II
12
0000
0043
0086 0128
0170
0212
0253
0294
0334
0374
4 8
13
12
17
16
21 25
20 24
30 34 38
28 32 37
0414
0792
0453
0828
0492
0864
0531
0899
0569
0934
0607
0969
0645
1004
0682
1038
0719
1072
0755
II06
4 8
4 7
3 7
3 7
12
II
II
10
15
IS
14
14
19 23
19 22
18 21
17 20
27 31 3S
26 30 33
25 28 32
24 27 31
13
14
15
l6
17
Is
19
20
II39
I46I
I173
1492
1206
1523
1239
ISS3
1271
1584
1303
1614
I33S
1644
1367
1673
1399
1703
1430
1732
3 7
3 I
3 6
3 6
10
10
9
9
13
12
12
12
16 20
16 19
IS 18
IS 17
23 26 30
22 25 29
21 24 28
20 23 26
I76I
1790
1818
1847
1875
1903
193 1
1959
1987
2014
3 6
3 5
9
8
II
II
14 17
14 16
20 23 26
19 22 25
2041
2304
2068
2330
2095
2355
2122
2380
2148
2405
2175
2430
2201
2455
2227
2480
2253
2504
2279
2529
3 5
3 5
3 5
2 5
8
8
8
7
II
10
10
10
14 16
13 15
13 IS
12 15
19 22 24
18 21 23
18 20 23
17 19 22
2553
2788
2577
2810
2601
2833
2625
2856
2648
2878
2672
2900
2695
2923
2718
2945
2742
2967
2765
2989
2 5
2 5
2 4
2 4
7
7
7
6
9
9
12 14
II 14
II 13
II 13
16 19 21
16 18 21
16 18 20
IS 17 19
3010
3032
3054
3075
3096
3118
3139
3160
3I8I
3201
2 4
6
8
II 13
IS 17 19
21
22
23
24
^1
11
29
30
31
32
33
34
II
11
39
3222
3424
3617
3243
3263
3464
365s
3284
3483
3674
3304
3502
3692
3324
3522
3711
3345
3541
3729
3365
3560
3747
338s
3579
3766
3404
3598
3784
2 4
2 4
2 4
6
6
6
8
8
7
10 12
10 12
9 II
14 16 18
14 15 17
13 IS 17
3802
3979
41SO
3820
3997
4166
3838
4014
4183
3856
4031
4200
3874
4048
4216
3892
4065
4232
3909
4082
4249
3927
4099
4265
3945
4116
4281
3962
4133
4298
2 4
2 3
2 3
5
5
S
7
7
7
9 II
9 10
8 10
12 14 16
12 14 IS
II 13 15
4314
4472
4624
4330
4487
4639
4346
4502
4654
4362
4518
4669
4378
4533
4683
4393
4548
4698
4409
4564
4713
4425
4579
4728
4440
4594
4742
4456
4609
4757
2 3
2 3
I 3
5
5
4
6
6
6
8 9
8 9
7 9
II 13 14
II 12 14 '
10 12 13
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
I 3
4
6
7 9
10 II 13
4914
5051
SI8S
4928
5065
5198
4942
S079
5211
4955
5092
5224
4969
5105
5237
4983
5119
5250
4997
5132
5263
SOU
5145
5276
5024
5159
5289
5038
5172
5302
13
I 3
I 3
4
4
4
6
S
5
7 8
6 8
10 II 12
9 II 12
9 10 12
S3IS
S44I
5563
5328
5453
5575
5340
5465
5587
5353
5478
5599
5366
5490
561I
537.8
5502
5623
5391
5514
563s
5403
5527
5647
5416
5539
5658
5428
5551
5670
I 3
I 2
I 2
4
4
4
5
5
5
6 8
6 7
9 10 II
9 10 II
8 10 II
5682
5798
591 1
5694
S809
S922
5705
S821
5933
5717
5832
5944
5729
5843
5955
5740
5855
5966
5752
5866
5977
5763
5877
5988
5775
5888
5999
5786
5899
6010
I 2
I 2
I 2
3
3
3
5
5
4
6 7
6 7
S 7
8 9 10
8 9 10
8 9 10
40
41
42
43
6021
6031
6042
6053
6064
6075
608 s
6096
6107
6117
I 2
3
4
S 6
8 9 10
6128
6232
633s
5138
5243
634s
6149
6253
6355
6160
6263
6365
6170
6274
6375
6180
6284
6385
6191
6294
6395
6201
6304
6405
6212
6314
6415
6222
632s
6425
I 2
I 2
I 2
3
3
3
4
4
4
5 6
789
789
789
44
643s
6532
6628
6444
6454
6551
6646
6464
6561
6656
6474
6571
6665
6484
6580
6675
6493
6590
6684
6503
6693
6513
6609
6702
6522
6618
6712
I 2
I 2
I 2
3
3
3
4
4
4
S 6
789
7 8 9
7 7 8
49
6721
6812
6903
6730
6821
6911
6739
6830
6920
6749
6839
6928
6758
6848
69J7
6767
6857
6946
677616785
6866 6875
6955 6964
6794
6884
6972
6803
6893
6981
I 2
1 2
1 2
3
3
3
4
4
4
5 5
4 5
4 5
678
678
678
50
6990
6998
7007
7016
7024
7033
7042 7OS0I7059
7067
I 2
3
3
4 sl 6 7 8
REVIEW OF SECONDARY SCHOOL ALGEBRA 477
LoGAEITHMS
lo ii|2 1 3(4ISl6|7\8 I9I123U 5 617 89I
51
52
53
7076:7084
7160 7168
72437251
7093
7177
7259
7101 7110
718s 7193
7267 7275
7118
7202
7284
7126
7210
7292
7135
7218
7300
7143
7226
7308
7152
7235
7316
I 2 3
12 2
12 2
3 4 5
3 4 5
3 4 5
678
677
6 6
54
55
73247332
7404 7412
7340
7419
7348 7356
7427 7435
7364
7443
7372
7451
7380
7459
7388
7466
7396
7474
12 2
12 2
3 4 5
3 4 5
667
5 6 7
56
7482 7490
7SS97S66
7634 7642
7497
7574
7649
7505 7513
7582 7589
7657 7664
7520
7597
7672
7528
7604
7679
7536
7612
7686
7543
7619
7694
7551
7627
7701
12 2
12 2
112
3 4 5
3 4 5
3 4 4
5 6 7
5 6 7
567
61
62
64
7709 7716
7782 7789
7853 7860
7723
7796
7868
7731 7738
7803 7810
7875 7882
7745
7818
7889
7752
782s
7896
7760
7832
7903
7767
7839
7910
7774
7846
7917
112
112
112
3 4 4
3 4 4
3 4 4
5 6 7
566
S 6 6
7924
7993
8062
7931
8000
8069
7938
8007
8075
7945 7952
80I4'802I
8082 8089
7959
8028
8096
7966 7973
8035 8041
8l02;8l09
7980
804S
8116
.7987
8055
8122
I Z 2
112
112
3 3 4
3 3 4
3 3 4
566
556
5 S 6
6s
8129
8136
8142
8149 8156
8162
8169
8176
8182
8189
112
3 3 4
5 56
66
?
69
70
71
72
73
74
8I9S
8261
832s
8202
8267
8331
8209
8274
8338
8215
8280
8344
8222
8287
8351
8228
8293
8357
8235
8299
8363
8241
8306
8370
8248
8312
8376
82S4
8319
8382
112
112
112
3 3 4
3 3 4
3 3 4
5 5 6
5 5 6
456
8388
84s I
8513
8395
8457
8519
8401
8463
8525
8407
8470
8531
8414
8476
8537
8420
8482
8543
8426
8488
8549
8432
8494
8555
8439
Ssoo
8561
8445
8506
8567
112
112
112
234
234
234
4 5 6
456
4 5 5
5f"
8633
8692
8579
8639
8698
!l*5
8645
8704
8591
8651
8710
till
8716
8603
8663
8722
8609
8669
8727
8615
867s
8733
8621
8681
8739
8627
8686
8745
Z I 2
112
112
234
234
234
4 5 5
4 5 5
4 5 5
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
112
233
4 5 5
76
11
8808
8865
8921
8814
8871
8927
8820
8876
8932
882s
8882
8938
8831
8887
8943
8837
8893
8949
8842
8899
8954
8848
8904
8960
8854
8910
8965
8859
8915
8971
112
112
112
233
233
233
4 5 5
4 4 5
4 4 5
81
83
84
8976
9031
908s
8982
9036
9090
8987
9042
9096
8993
9047
910I
8998
9053
9106
9004
9058
9112
9009 901S
9063 '9069
9117J9122
9020
9074
9128
902s
9079
9133
112
112
112
233
233
233
4 4 5
4 4 5
4 4 5
9138
9I9I
9243
9143
9196
9248
9149
9201
9253
9154
9206
9258
9159
9212
9263
9165
9217
9269
9170
9222
9274
9175
9227
9279
9180
9232
9284
9186
9238
9289
112
112
112
233
233
233
4 4 5
4 4 5
4 4 5
85
86
9294 9299
9304
9309
93IS
9320
9325
9330
9335
9340
112
233
4 4 5
9345
939S
9445
9350
9400
9450
9355
940s
9455
9360
9410
9460
9365
9415
946s
9370
9420
9469
9375
9425
9474
9380
9430
9479
9385
9435
9484
9390
9440
9489
112
Oil
0 1 1
233
223
223
4 4 5
3 4 4
3 4 4
89
90
91
9494
9542
9590
9499
9547
9595
9504
9552
9600
9509
9SS7
960s
9513
9562
9609
9518
9566
9614
9523 9528
9571 9576
9619 9624
9533
9581
9628
9538
9586
9633
Oil
0 1 1
Oil
223
223
223
3 4 4
3 4 4
3 4 4
92
93
94
9638
968s
9731
9643
9689
9736
9647
9694
9741
9652
9699
9745
9657
9703
9750
9661
9708
97S4
9666 9671
9713 9717
9759 9763
9675
9722
9768
9680
9727
9773
Oil
0 1 1
0 1 I
223
223
223
3 4 4
3 4 4
3 4 4
95
96
9777
9782
9786
9791
9795
9800
980s 9809
9814
9818
Oil
223
3 4 4
9823
9868
9912
9827
9872
9917
9832
9877
9921
9836
9881
9926
9841
9886
9930
984s
9890
9934
9850 9854
9894 9899
9939 9943
9859
9903
9948
9863
9908
9952
0 1 I
oil
0 1 I
223
223
223
3 4 4
3 4 4
3 4 4
99
9956
9961
9965
9969
9974
9978
9983 9987
9991
9996
oil
223
3 3 4
The copyright of that portion of the above table which gives the logarithms of
numbers from 1000 to 2000 is the property of Messrs. Macmillan and Company,
limited, who, however, have authorised the use of the form in any reprint pub-
lished for educational purposes.
478 ELEMENTARY MATHEMATICAL ANALYSIS
Logarithms of Tbiqonometric Fttnctions
o /
log sin
d
log tan
dc
log cot
log cos
'
S
T
0.0000
0 90
0
0 w
10
7.4637
3011
7-4637
3011
2.5363
0.0000
so
10
6.4637
6.4637
20
7.7648
1760
12S0
969
7.7648
1 761
1249
969
2.2352
0.0000
40
20
6.4637
6.4637
30
7.9408
7.9409
2.0591
0.0000
30
30
6.4637
6.4637
40
8.0658
8.0658
1.9342
0 . 0000
20
40
6.4637
6.4637
SO
8.1627
792
669
580
8.1627
792
670
s8o
I . 8373
0.000b
10
50
6.4637
6.4638
I 0
8.2419
8.2419
1.7581
9.9999
0 89
60
6.4637
6.4638
10
8.3088
8 . 3089
1.6911
9.9999
SO
70
6.4637
6.4638
20
8.3668
SII
458
413
8.3669
5"
457
41S
1.6331
9.9999
40
80
6.4637
6.4638
30
8.4179
8.41S1
I.S8I9
9.9999
30
90
6.4637
6.4638
40
8.4637
8 . 4638
1.5362
9.9998
20
100
6.4637
6.4638
so
8.S0S0
378
348
321
8.S053
378
348
322
1.4947
1. 4569
9.9998
10
110
6.4637
6.4639
2 0
8.S428
8.S43I
9.9997
0 88
120
6 . 4636
6.4639
10
8.S776
8.5779
1.4221
9.9997
SO
130
6.4636
6.4639
20
8 . 6097
300
280
8.6101
300
281
263
1.3899
9.9996
40
140
6 . 4636
6.4640
30
8.5397
8.6401
8.6682
I.3S99
9.9996
30
ISO
6 . 4636
6.4640
40
8.6677
263
1.3318
99995
20
160
6.46J6
6 . 4640
so
8 . 6940
248
235
222
8.694s
249
235
223
1.3OSS
9.9995
10
170
6.463s
6.4641
3 0
8.7188
8.7194
1.2806
9.9994
0 87
180
6.463s
6.4641
10
8.7423
8.7429
I.2S71
9.9993
SO
190
6.4635
6.4642
20
8.764s
212
8.7652
213
1 . 2348
9.9993
40
200
6.463s
6.464a
30
8.7857
8.786s
I. 213s
9.9992
30
210
6.463s
6.4643
40
8.8059
192
8 . 8067
194
1.1933
9.9991
20
220
6.4634
6.4643
so
8.82SI
18s
177
170
8.8261
178
171
1.1739
9.9990
10
230
6.4634
6.4644
4 0
8 . 8436
8.8613
8 . 8446
8.8624
I.1SS4
9.9989
0 86
240
6.4634
6.4644
10
1.1376
9.9989
SO
250
6.4633
6.4645
20
8.8783
IS8
152
8.879s
IS8
154
1.1205
9.9988
40
260
6.4633
6 . 4646
30
8.8946
8. 8960
1 . 1040
9.9987
30
270
6.4633
6 . 4646
40
8.9104
8.9118
1.0882
9.9986
20
280
6.4632
6.4647
so
8.9256
147
8.9272
148
1.0728
9.998s
10
290
6.4632
6.4648
5
0
8.9403
8.9420
1.0580
9.9983
0 8s
300
6.4631
6 . 4649
1 log COB 1 d 1 log cot^ I dc 1 log tan
Hog Sin 1 ' ° 1 1 1 1
113
142
138
137
13S
134
130
129
19
7
12s 1
23
1S2
119
117
lis
lU
1
U.3
14.2
13.8
13.
7 13.5
13.4
13.
» 12.9
12
.7
12.6
12.3
12.2
11.
) 11.7
11.5
11.4
2
28.6
28.4
27.6
27.
4 27.0
26.8
26.
a 25.8
2S
.4
25.0
!4.6
24.4
23.
i 23.4
23.0
22.8
3
42.9
42.6
41.4
41.
1 40.5
40.2
39.
D 38.7
3E
.1
37.5
16.9
36.6
35.
J 35.1
34.5
34.2
1
57.2
56.8
55.2
54.
8 54.0
53.6
52.
9 51.6
6C
.8
50.0 '
19.2
48.8
47.
i 46.8
46.0
45. 6
S
71.5
71.0
69.0
68.
6 67.5
67.0
65.
a 64.5
63
.5
62.5
H.5
61.0
59.
> 58.6
67.6
57.0
6
85.8
85.2
82.8
82.
2 81.0
80.4
78.
D 77.4
7(
.2
75.0
r3.8
73.2
71.
I 70.2
69.0
68.4
7
100.1
99.4
96.6
95.
9 94.5
93.8
91.
D 90.3
8!
.9
87.6
36.1
85.4
83.
i 81.9
80.5
79.8
8
114.4
113.6
110.4
109.
6 108.0
107.2
104.
0 103.2
101
.6
100.0
98.4
97,6
95.
! 93.6
92.0
91.2
128.7
127.8
124.2
123.
3 121.5
120.6
117.
D 116.1
IM
.3
112. 5 1
10.7
109.8
107.
1105.3
103.5
102.6
Formulas for using
Table directly
% ! log sin * = log I* + S °io
log cos X = log (90 - *)' + S
log cot * = log (90 - *)' + T
log tan X = colog (90 — xy + co T
V log tan X = log I* + T "
« [ log cot * = colog I* + CO T ^
REVIEW OF SECONDARY SCHOOL ALGEBRA 479
Logarithms of Tbigonometeic Functions
log sin
log tan
dc
log cot
log COS
pp
so
so
SO
8.9403
8.9S4S
8. 9682
8.9816
8. 9945
9 . 0070
9.0192
9.031I
9.. 0426
9.0S39
9 . 0648
9.07SS
9.0859
9 . 0961
9.1060
9.1157
9.1252
9.1345
9.1436
9.1525
9.1612
9.1697
9.1781
9.1863
9-1943
9.2022
9.2100
9.2176
9.2251
9.2324
9.2397
log COS
142
137
134
129
125
122
119
IIS
113
109
107
104
102
99
97
95
93
9f
89
8t
85
84
82
80
79
78
76
75
73
8.9420
8.9563
8.9701
8.9836
8 . 9966
9.0093
9.0216
9.0336
9.0453
9.0567
9.0678
9.0786
9.0891
9.0995
9 . 1096
9.1194
9.1291
9.1385
9.1478
9.1569
0.1658
9. 1745
9.1S31
9.1915
9.1907
9.2078
9.2158
9.2236
9.2313
S-2389
9.2463
log cot
143
138
13s
130
127
123
120
117
114
111
108
105
104
lOI
98
97
94
93
91
89
87
86
84
82
81
80
78
77
76
1.0580
1.0437
1.0299
1.0164
1.0034
0.9907
0.9784
0.9664
0.9547
0.9433
0.9322
0.9214
0.9109
0.9005
0.8904
9.8806
0.8709
0.8615
0.8522
0.8431
o . 8342
0.8255
0.8169
0.8085
0.8003
0.7922
0.7842
0.7764
0.7687
0.7611
0.7537
9.9983
9.9982
9.9981
g.9980
9.9979
9.9977
9.9976
9.9975
9.9973
9.9972
9.9971
9.9969
9.9968
9.9966
9.9964
9.9963
9.9961
9.9959
9-9958
9.9956
9.9954
9.9952
9.9950
9.9948
9.9946
9.9944
9.9942
9.9940
9.9938
9.9936 10
9.9934 O 80
o 8S
10
0 84
50
40
30
20
o 83
40
30
20
10
0 82
SO
40
30
20
0 81
dc
log tan
log sin
"3
II. 3
22.6
33.9
III
II. I
22.2
33.3
45.2
67i8
44.4
55. S
66.6
79.1
90.4
101.7
77.7
88.8
99.9
108
10.8
21.6
32.4
107
10.7
21.4
32.1
43.2
64.8
42.8
53. 5
64.2
75.6
86.4
97-2
74.9
85.6
96.3
104
10.4
20.8
31.2
102
10.2
20.4
30.6
41.6
52.0
62.4
40.8
51.0
61.2
72.8
83.2
93.6
il'.6
91.0
109
10. 9
21.8
32-7
43-6
54-5
65.4
76.3
87.2
98.1
lOS
10.5
21.0
3I-S
42 -0
525
63.0
73. S
84.0
94-S
lOI
10. 1
20.2
30.3
40.4
50.5
60.6
70.7
80.8
90.9
94
9.4
18.8
28.2
37.6
47.0
58.4
65.8'
75.2
84.6!
18.6
27.9
37.2
46.9
SS.S
65.1
74.4
83.7
91
9.1
18.2
27.3
36.4
46.5
54. 6
63.7
72.8
81.9
89
8.1
17.8
26.7
35.6
44.5
53.4
62.
71.2
80.1
87
8.7
17.4
26.1
31.8
43.5
92.2
60.9
69.6
78.3
86
8.6
17.2
25.8
34.4
43.0
91.6
77.4
86
8.5
17.0
15.5
34.0
42.9
91.0
59.9
68.0
76.9
84
8.4
16.8
25
33
42.0
50.4
98.8
67.2
79.6
16.4
24.6
32.8
41.0
49.2
67.4
65.6
73.8
81
8.1
16.2
2.34
32.4
40
48.6
56.7
64.8
72.9
79 I 78
7.9 7.8
19.815.6
23.7,23.4
31.631.
39.9;39.
47.446.
0
8
95. 3^94. 6
63.262.4
71.1170.2
99
9.9
19-8
29-7
39.6
49. S
S9.4
69.3
79-2
89 -I
77
7.7
15.4
23.1
30.1
38.!
46.:
98
9.8
19.6
29.4
39.2
49
58.8
68.6
78.4
97
9-7
19.4
29.1
38.8
95
9.5
19.
28.5
38. 0
48.5 47. S
57. o
58.2
67.9
77-6
87.3
66. s
76.0
85. S
76 1 78 I 74
7.6 7.5 7.4
19.2:19.014.8
1.5
61
78
7.3
14.6
21.9
l.4'30.
1.037,
i.649
.4167
0'29. 629.2
537.0:36.5
.0j44.4p.g
ffsi.ffgi.i
.099.298.4
.9|68.6i65.7
Formulas for usine Table inversely
Ilog *'■= log sin X — S
log «' = log tan X — T
colog a/ = log cot X — CO T
log (90 — x)' '
log (90 - x)' •
colog (90 — xy
' log
■■ log
■ log
COS
cot
tan
X —
X —
S
T
CO T
480 ELEMENTARY MATHEMATICAL ANALYSIS
Logarithms op Trigonometbic Functions
o /
log sin
d
log tan
dc
log cot
log cos
d
pp
10 0
9 2397
71
70
9-2463
0-7S37
9.9934
3
2
0
80
73
71
10
9.246S
9-2536
73
73
0.7464
9.9931
SO
I
2
7-3
14.6
7-1
14.2
20
9.2538
68
9-2609
71
70
69
0.7391
9-9929
2
40
3
21.9
21-3
30
40
9 . 2606
9.2674
68
66
9-2680
9.2750
0.7320
O.72SO
9-9927
9.9924
3
2
30
20
4
5
29.2
36. 5
28.4
35. S
50
9.2740
66
9.2819
68
0.7181
9.9922
3
2
10
6
43.8
42.6
II 0
9.2806
54
64
9.2887
66
67
0.7113
9.9919
0
79
7
SI.l
4?Z
10
9.2870
9.2953
0.7047
9.9917
3
50
8
9
S8.4
6S.7
S6.8
63.9
20
9.2934
53
61
61
9.3020
65
63
0.6980
9-9914
2
40
70
7.0
14.0
59
6.9
13.8
30
9.2997
9.3085
0.691S
9.9912
3
2
30
40
9-3058
9.3149
0.6851
9.9909
20
2
so
9.3119
60
59
58
9.3212
63
61
61
0.6788
9-9907
3
3 .
2
10
3
21.0
20.7
12 0
9.3179
9.3275
0.672s
9.9904
0
78
4
28.0
27.6
10
9.3238
9-3336
0 . 6664
9-9901
SO
35.0
42.0
34-S
41.4
20
9.3296
57
9.3397
61
0 . 6603
9.9899
3
3
3
40
7
8
9
49.0
56.0
63.0
48-3
55-2
62.1
30
40
9.3353
9.3410
9.3458
9-3517
59
59
0.6542
0.6483
9.9896
9.9893
30
20
SO
9.3466
55
54
54
9-3576
58
57
57
0.6424
9.9890
3
3
3
10
«
68
67
13 0
9.3521
9-3634
0.6366
9.9887
0
77
I
6.8
6.7
10
9.357s
9-3691
0.6309
9.9884
50
2
3
13.6
20.4
13.4
20. I
20
9.3629
53
S2
52
9-3748
56
55
5S
0.6252
9.9881
3
3
3
40
4
5
6
27.2
34-0
40.8
26.8
30
9.3682
9.3804
0.6196
9.9878
30
33.5
40.2
40
9-3734
9.3859
0.6141
9.987s
20
SO
9-3786
SI
SO
50
9.3914
54
53
S3
0.6086
9.9872
3
3
3
10
7
47.6
46.9
14 0
9.3837
9.3968
0.6032
9.9869
0
76
8
54.4
53-6
ID
9.3887
9.4021
0.5979
9 . 9866
50
9
61.2
60.3
20
9.3937
49
48
9.4074
S3
SI
52
0.5926
9.9863
4
3
3
40
66
6.6
6.5
13.0
19. 5
30
9-3986
9-4127
0.5873
9.9859
30
13.2
19-8
40
9 -403s
9.4178
0.5822
9.9856
20
3
SO
9.4083
47
9.4230
51
0.5770
9. 9853
4
10
4
26..^
26.0
IS 0
9-4130
9.4281
0.S7I9
9.9849
0
75
5
33-0
32.5
6
7
39-6
46.2
39.0
45.5
log COS
d
log cot
dc
log tan
log sin
d
°
8
9
52.8
59.4
SO
52.0
58.5
48 1 47
64
fS
6
I
6
0
59
S8
57
S6,
55
54
53
52
31 1
I
6.4
6.3
6
.1
6
.0
S.9
5.8
5.7
5.6
S-S
5-4
5-3
s
2
S.I
5.0
4-8! 4.7
2
12.8
12.6
12
.2
12
.0
11.8
II. 6
II-4
II. 2
I.O
10.8
10.6
TO
4
10.2 1
0.0
9.6 9.4
3
19.2
18.9
18
■ 3
18
.0
17.7
17-4
17-I
16.8
6.S
16.2
15-9
15
6
IS. 3 I
5.0 I
4-4
14.1
4
2S.6
2S.2
24
•4
24
.0
23.6
23-2
22.8
22.4 .
>2.0
21.6
21-2
20
8
20.4!
0.0 )
9-2
18.8
S
32.0
3I-S
30
■ 5
30
.0:29.5
29.0
28. s
28.0 i
7-S
27.0
26.5
26
0
25.5:
S.o:
4.0
23-5
6
38.4
37.8
36
.6
36
-035-4
34-8
34-2
33.6.
i3-0
32.4
31.8
31
2
30.6 :
0.0 i
8.8
28.2
7
44.8
44.1
42
.7
42
.041.3
40-6
39-9
39-2 ,
i8-5
37-8
37-1
36
i
3S.7:
S-o;
3-6
32.9
8
SI. 2
SO. 4
48
.8
48
.047.2
46-4
45-6
44.8-
t4-0
43-2
42-4
41
40.8 i
io-o ;
8.4
37.6
9
S7.6
S6.7
54
-9
54
.o;s3.i
S2.2
SI-3
50.4 '
J9-S
48-6
47.7
46
8
45.9^
tS.Oi
J3-2
42.3
KEViEW OF SECONDARY SCHOOL ALGEBRA 481
Logarithms op Thigonomethic Functions
log sin
log tan
dc
log cot
log COS
IS o
20
30
40
i6 0
i8 0
20
30
40
19 o
9.4130
9.4177
9.4223
9.4269
9.4314
9.43S9
9.4403
9-4447
9-4491
9-4533
9.4576
9.4618
9.4659
9.4700
9-4741
9.4781
9.4821
9.4861
9.4900
9-4939
9.4977
9-5015
9-S052
9.S090
9.5126
9.S163
9.5199
9-5235
9.5270
9-^306
9-5341
9.4281
9.4331
9.4381
9.4430
9.4479
9.4527
9-4575
9.4622
9.4669
9-4716
9-4762
,4808
4853
14898
9-4943
9-4987
9-5031
9-5075
9-5118
9.5161
9.5203
9-5245
9-5287
9-5329
9-5370
9-5411
9-5451
9-5491
9-5531
9-5571
9-S611
0-5719
0-5669
0,5619
0-5570
0.5521
0-5473
0-5425
0.5378
0-5331
0.5284
O-S238
0.5192
0-S147
0.5102
0-5057
0.5013
o . 4969
0.492s
0.4882
0.4839
0.4797
0.475s
0.4713
0.4671
0.4630
0.4589
0.4549
0.4509
0.4469
0.4429
0.4389
9-9849
9.9846
9.9843
9.9839
9.9836
9.9832
9.9828
9.9825
9.9821
9.9817
9-9814
9-9S10
9.9806
9.9802
9.9798
9-9794
9.9790
9.9786
9-9782
9.9778
9.9774
9.9770
9-9765
9.9761
9.9757
9.9752
9-9748
9 9743
9.9739
9-9734
9-9730
0 75
0 72
50 71
25.0
30.0
35-0
40.0
45 -Q
48
4.8
9.6
14.4
ig.2
24.0
28.8
33.6
38.4
43
49
4-9
9.8
14-7
log cos
log cot
dc
log tan
log sin
46 1
4
6
0
2
13
8
18
4
2,1
0
27
6
,12
2
36
8
41
4
47
4-7
9-4
14-1
18.8
23. 5
28-2
32.9
37-6
42-3
45
4-5
9.0
13.5
31.5
36.0
40,5
44
13-2
43
12.9
42
4,2
8-4
12-6
1
2
3
41
12.3
40
12.0
39
7.8
1 1.. 7
3.8
7.6
11.4
1
2
3
37
3-7
7-4
11- 1
17-6
22.0
26.4
17.2
21. 5
2S.8
16.8
21.0
25.2
4
5
6
16.4
20.5
24.6
16.0
20.0
24.0
15.6
19. 5
23-4
15.2
19.0
22.8
4
5
6
14.8
18.5
22.2
30.8
35.2
39.6
30.1
34-4
38.7
29.4
33-6
37-8
7
8
9
28.7
32.8
36.9
28.0
32.0
36.0
27-3
31-2
3S-I
26.6
30.4
34-2
7
8
9
25-9
29.6
33.3
36
3.6
7.2
10.8
14-4
18-O
21.6
25.2
28.8
32.4
35
35
7-0
10-5
14.0
17-5
24- 5
28.0
31-5
31
482 ELEMENTARY MATHEMATICAL ANALYSIS
LOQABITHMS OF TbIOONOMETRIC FUNCTIONS
■ 1
log Sin
d
log tan
dc
log cot
log COS
d
PP
ao 0
9.S34I
34
34
9.5611
39
39
0.4389
9.9730
S
4
0 70
10
9-S37S
9.5650
0.4350
9.9725
50
4
0.4
20
9.S409
34
34
33
9.5689
38
38
0.43"
9,9721
5
40
I
30
40
9-S443
9.S477
0.5727
9.5766
0.4273
0.4234
9.9716
9.9711
5
5
30
20
2
3
0.8
1.2
so
9-SSiO
33
33
33
9.5804
38
38
0 . 4196
9.9706
4
5
5
10
4
1.6
31 0
9.SS43
9.5842
0.4158
9.9702
0 69
5
2,0
10
9.SS76
9.5879
0.4121
9.9697
SO
6
2.4
30
9.S609
32
32
31
9.S9I7
37
37
37
0.4083
9.9692
5
5
5
40
7
2.8
30
9.S641
9.5954
0.4046
9.9687
30
8
3.2
40
9.5673
9.5991
0.4009
9.9682
20
9
3.6
so
9.S704
32
31
31
9.6028
35
36
0.3972
9.9677
5
5
6
10
aa 0
9.5736
9.6064
0.3936
9.9672
0 68
10
9.5767
9.6100
0.3900
9.9667
SO
I
2
5
0.5
1.0
20
9.5798
30
31
30
9.6136
36
36
35
0.3864
9.9661
5
S
5
40
30
9.5828
9.6172
0.3828
9.9656
30
3
1.5
40
9.5859
9.6208
0.3792
9.9651
20
4
5
2.0
2.5
SO
9.5889
30
29
30
9.6243
36
35
34
0.3757
9.9646
6
10
23 o
9.5919
9.6279
0.3721
9.9640
%
0 67
6
3.0
10
9.5948
9.6314
0.3686
9.9635
50
7
3.5
20
9.5978
29
29
29
9.6348
35
34
35
0.3652
9.9629
1
40
8
4.0
30
9.6007
9.6383
0.3617
9.9624
30
9
4.5
40
9.6036
9.6417
0.3583
9.961:8
5
20
so
9.6065
28
9.6452
34
34
33
0.3548
9.9613
6
10
24 0
9.6093
28
9.6486
0.3514
9.9607
%
0 66
6
10
9.6121
28
9.6520
0.3480
9.9602
SO
I
0.6
20
9.6149
28
9.6SS3
34
33
34
0.3447
9.9596
6
40
2
3
I . 2
1.8
30
9.6177
28
9.6587
0.3413
9.9S90
5
30
40
9.6205
27
9.6620
0.3380
9.9584
5
20
4
2.4
so
9.6232
27
9.6654
33
0.3346
9.9579
6
10
\
3.0
3.6
as 0
9.6259
9.6687
0.3313
9.9573
0 6s
9
4.8
5.4
log cos
d
log cot
dc
log tan
log sin
d
/ 0
39
38„
37
36
35
34
33
32
31
30
29
38
27
I
3.9
3-8
3-7
3.6
3.5
3.4
3.3
3.2
3.
I 3.0
2.9
2.8
2.7
2
7.8
7.6
7.4
7.2
7.0
6.8
6.e
6.4
6.
2 6.0
S.8
5.6
5.4
3
II. 7
II. 4
II. I
10.8
10. 5
10.2
9.S
9.6
9.
3 9.0
8.7
8.4
8.1
4
15.6
IS. 2
14.8
14.4
14.0
13.6
13.!
12.8
12.
\ 12.0
11.6
II. 2
10.8
S
19. 5
19.0
18.5
18.0
17.5
17.0
16. s
16.0
IS.
5 iS.o
14. S
14.0 13. 5
16.8 16.2
6
23.4
22.8
22.2
21.6
21.0
20.4
19. 8
19.2
18.
3 18.0
17.4
7
27.3
26.6
25.9
25.2
24.5
23.8
23.1
22.4
21.
^ 21.0
20.3
19.6 18.9
8
31.2
30.4
29.6
28.8
28.0
27.2
26. j{
25.6
24.
5 24.0
23.2
22.4 21.6
9
35. 1
34-2
33-3
32.4
31. 5
30.6
29.7
28.8
27.
J 27. 026. 1
25.2 24.3
REVIEW OF SECONDARY SCHOOL ALGEBRA 483
Logarithms of Thigonometbic Functions
log Bin d log tan
dc log cot log cos d
iS 0
9.62S9
9.6286
9.6313
30 9.6340
40 9 . 6366
26 o
SO
so
28 0
10
30 0
9.6392
9.641S
9.6444
9.6470
9.649s
9.6S21
9.6546
9.6370
9.6595
9.6620
9.6644
9.6668
9 . 6692
9.6716
9.6740
9.6763
9.6787
9.6810
9.6833
9.6856
9.6878
9.6901
9.6923
9.6946
9.6968
9.6990
log cos
9.6687
9.6720
9.6752
9.6785
9.6817
9.6850
9.6882
9.6914
9.6946
9.6977
9 . 7009
9.7040
9.7072
9.7103
9.7134
9.7165
9.7196
9.7226
9.7257
9.7287
9.7317
9.7348
9.7378
9.7408
9-7438
9.7467
9.7497
9.7S26
9.7356
9.758s
9.7614
log cot
dc
0.3313
0.3280
0.3248
0.321S
0.3183
0.3150
0.3118
0.3086
0.3054
0.3023
0.2991
o . 2960
0.292&
0.2897
0.2866
0.2835
0.2804
0.2774
0.2743
0.2713
0.2683
0.2652
0.2622
0.2392
0.2562
0.2533
0.2503
0.2474
0.2444
0.2415
0.2386
log tan
9.9573
9.9567
9.9361
9.9SS3
9-9549
9-9543
9-9S37
9-9530
9-9524
9-9518
9-9512
9-9505
9-9499
9.9492
9.9486
9.9479
0-9473
9 - 9466
9-9459
9-9453
9.9446
9.9439
9.9432
,9425
.9418
.9411
9-9404
9-9397
9-9390
9-9383
9-937S
log sin
6S
o 64
0 63
SO
o 62
o 61
o 60
6
0.6
1.2
1.8
2.4
3.0
3.6
7
0.7
1.4
2.1
2.8
3.5
4-2
4-9
3-6
6-3
8
0.8
1.6
2.4
3.2
4-0
4-8
3-6
6-4
7-2
,13
32
31
30
29
27
26
2S
24
23
T
3-3
3-2
3.1
I
3-0
2.0
2.7
2.6
2-5
I
2.4
2.3
2
6.6
6-4
6.2
2
6.0
S.8
5-4
S-2
5-0
2
4-8
4-6
3
9-9
9-6
9.3
3
9-0
8.7
8-1
7.8
7-5
3
7-2
6-9
4
13-2
12.8
12.4
4
12.0
II. 6
10-8
10.4
10 -0
4
9-6
9-2
S
I6-I!
16.0
IS. 5
5
15-0
14-S
13-5
13.0
12-3
S
12.0
ir.s
6
19-8
19.2
18.6
6
18.0
17.4
16.2
IS-6
15.0
6
14-4
13.8
7
23.1
22.4
25.6
21.7
'7
21.0
20.3
18-9
18.2
17-5
7
16-8
16. 1
8
26.4
24.8
8
24.0
23.2
21.6
20.8
20.0
8
19-2
1S.4
9
29-7
28.8
27.9
9
27.0
26.1
24-3
23-4
22.5
9
21-6
20.7
4-4
6.6
8.8
15-4
17.6
19.8
484 ELEMENTARY MATHEMATICAL ANALYSIS
LOGABITHMS OP TSiaONOMBTKIC FUNCTIONS
log sin
log tan
dc
log cot
log COS
30
0
10
9.6990
9.7012
20
30
40
9.7033
9.70SS
9.7076
31
SO
0
10
9.7097
9.7II8
9.7139
20
30
40
9.7160
9.7I8I
9.7201
32
so
0
10
9.7222
9.7242
9.7262
20
30
40
9.7282
9.7302
9.7322
33
so
0
10
9.7342
9.7361
9.7380
4
20
30
40
9.7400
9.7419
9.7438
34
so
0
10
9.74S7
9.7476
9.7494
20
30
40
9.7SI3
9.7S3I
9.7SSO
35
so
0
9.7S68
9.7586
9.7614
9.7644
9.7673
9.7701
9.7730
9.77S9
9.7788
9.7816
9.7845
9.7873
9 . 7902
9.7930
9.7958
9.7986
9 . 8014
9 . 8042
9.S070
9.8097
9.812s
9.8153
9.8180
9.8208
9.823s
9 . 8263
9.8290
9.8317
9.8344
9.8371
9.8398
9.8425
9.8452
0.2386
0.2356
0.2327
0.2299
0.2270
0.2241
0.2212
0.2184
O.21SS
0.2127
0.2098
0.2070
0.2042
0.2014
0.1986
O.I9S8
0.1930
0.1903
0.1875
0.1847
0.1820
0.1792
0.176s
0.1737
0.1710
0.1683
o.i6s6
0.1629
0.1602
0.1575
0.1548
9.937s
9.9368
9.9361
9.9353
9.9346
9.9338
9.9331
9.9323
9.931s
9.9308
9.9300
9.9292
9.9284
9.9276
9.9268
9.9260
9.9252
9.9244
9.9236
9.9228
9.9219
9.9211
9.9203
9.9194
9.9x86
9.9177
9.9169
9.9160
9.9151
9.9142
9.9134
o 60
o 58
0 56
log COS
log cot
log tan
log sin
30
3.0
6-0
9.0
29
2.9
S.8
8.7
12.0
11.6
15.0
18.0
14-5
17.4
21.0
20.3
24.0
27.0
23.2
26.1
2S
2.8
5.6
8.4
11.2
14.0
16.8
19.6
4
25.2
27
22
I
2.7
2.2
2
3
5.4
8.1
Vi
4
10.8
8.8
i
13. S
16.2
11.0
13.2
7
8
18.9
21.6
IS. 4
17.6
9
24.3
19.8
4.2
6.3
8.4
0.5
14-7
16.8
18.9
20
10
I
2.0
1.0
2
4.0
.1.8
3
6.0
5.7
4
8.0
7.6
S
10. 0
9.5
6
12.0
11.4
7
14.0
16.0
13.3
8
15.2
9
18.0
17. 1
7
0.7
1.4
2.8
3.5
4.2
4.9
S.6
6.3
8
0.8
1.6
2.4
3.2
4.0
4.8
5.6
6.4
7.2
9
0.9
18
1.8
3.6
5.4
7.2
9.0
10.8
12.6
14.4
16.2
REVIEW OF SECONDARY SCHOOL ALGEBRA 485
LOQAEITHMS OF TeIGONOMBTBIC FUNCTIONS,
log sin
log tan
dc
log cot
log cos
3S 0
30
40
so
36 o
30
40
so
37 0
10
20
30
40
SO
38 o
39 o
9-7585
9 . 7604
9.7622
9.7640
9.7657
9-7675
9.7692
9.7710
\
9.7727
9.7744
9.7761
9.7778
9-7795
9-7811
9.7S28
9-7844
9.7861
9-7877
9.7893
9.7910
9.7926
9.7941
9.7957
9.7973
9.7989
9.8004
9.8020
9.8035
9.8050
9.8066
9.8081
log COS
9.8452
9.8479
9.8506
9.8533
9.8559
9-8586
9-8613
9-8639
9.8666
9.8692
9.8718
9.8745
9.8771
9.8797
9.8824
9-8850
9-8876
9.8902
9.8928
9.8954
g.8980
9 . 9006
9.9032
9-9058
9.9084
9.9110
9.9135
9.9161
9.9187
9.9212
9.9238
log cot
dc
0.1548
O.IS2I
0.1494
o . 1467
O.I44I
O.I414
0.1387
O.I36I
0.1334
0.1308
0.1282
O.I2S5
0.1229
0.1203
O.II76
Q.I150
o. 1124
0.1098
0.1072
0.1046
0.1020
0 . 0994
0.0968
0.0942
0.0916
0.0890
0.086s
0.0839
0.0813
p. 0788
0.0762
9-9134
9-9125
9-9I16
9-9107
9.9098
9.9089
9.9080
9.9070
9.9061
9-9052
9 . 9042
9-9033
9-9023
9-9014
9.9004
9.8995
9-8985
9-8975
9-8965
9-8955
9-8945
9-8935
9-8925
9-891S
9-8905
9-8895
9.8884
9-8874
9 - 8864
9-8853
9.8843
log tan
log sin
o 55
0 S4
50
40
30
20
o S3
50
9
0.9
1.8
2-7
3-6
4-5
5-4
6-3
7-2
8.1
4.4
S-5
6-6
7.7
8.8
9-9
27 1
2
7
s
4
8
I
10
8
1.1
5
16
2
t8
9
21
6
24
3
26
2.6
S.2
7.8
:o.4
13 o
IS. 6
18.2
20.8
23-4
Z-S
18
2-5
1.8
S-0
3.6
7-5
5.4
10. 0
7-2
r2.5
9-0
15.0
10.8
17.5
12.6
20.0
14.4
22.5
16.2
17
1-7
3-4
5-1
6.8
8-5
II. 9
13-6
IS-3
16
1.6
3.2
4-8
6-4
8-0
9-6
IS
l-S
3-0
4-5
6.0
7-S
9-0
486 ELEMENTARY MATHEMATICAL ANALYSIS
LOQARITHMS OF TRIGONOMETRIC FUNCTIONS
log Bin
log tan
dc
log cot
log COS
42 o
43 o
44 o
50
45 0
9 . 8081
9 . 8096
9.8111
9.812s
9.8140
9.81SS
9.8169
9.8184
9.8198
9.8213
9.8227
9.8241
9.82SS
9.8269
9.8283
9.8297
9.8311
9.8324
9.8338
9.8351
9.836s
9-8378
9.8391
9.840s
9.8418
9.8431
9.8444
9.84S7
9 . 8469
9.8482
9.849s
9.9238
9.9264
9.9289
9.931S
9.9341
9 . 9366
9.9392
9.94*7
9.9443
9.9468
9.9494
9.9SI9
9. 9544
9.9S70
9.9S9S
9. 9621
9.9646
9.9671
9.9697
9.9722
9.9747
9.9772
9.9798
9.9823
9.9S48
9.9874
9.9899
9.9924
9.9949
9. 9975
O . 0000
0.0762
0.0736
0.07II
0.068s
o.o6s9
0.0634
o . 0608
0.0583
0.0SS7
O.OS32 ,
0.0506
0.0481
0.0456
0 . 0430
0 . 040s
0.0379
0.0354
o . 0329
0 . 0303
0.0278
0.0253
0.0228
0.0202
0.0177
0.0IS2
0.0126
O.OIOI
0.0076
0.0051
0.002s
0 . 0000
log COS
log cot
dc
log tan
9.8843
9.8832
9.8821
9.88IC
9.8800
9.8789
9.8778
9.8767
9.8756
9.8745
9.8733
g.8722
9.87II
9 . 8699
9.8688
9.8676
9 . 8665
9.8653
9.8641
9.8629
0.8618
9.8606
9.8594
9.8582
9.8569
9.8557
9.8545
0.8532
9.8520
9.8507
9.8495
0 49
so
o 48
SO
40
30
0 47
50
40
30
20
0 46
SO
40
30
20
log sin
26
2.6
S.2
7.8
10.4
13.0
IS. 6
18.2
20.8
23.4
25
2.5
so
.7.5
10. o
12. S
IS.O
17. s
20.0
22. S
15
1. 5
3.0
4.5
6.0
7.5
9.0
10. s
12.0
13. 5
14
1.4
2.8
4.2
5.6
7.0
8.4
9.8
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
13
1.3
2.6
3.9
5.2
6.5
7.8
9.1
10.4
II. 7
4.4
s.s
6.6
7.7
8.8
9.9
12
1.2
1:1
4.8
6.0
7.2
■8.4
9.6
10.8
REVIEW OF SECONDARY SCHOOL ALGEBRA 487
Natural Trigonometric Functions
Deg.
Radians
n Bin
n CSC
n tan
n cot
n sec
n cos
o
0 . 0000
.000
.000
1. 000
1. 00
1.5708
90
I
2
3
0.0I7S
0.0349
0.0524
.017
.035
.052
57.3
28.7
19. 1
.017
.035
.052
57.3
28.6
19. 1
1. 000
1. 001
1. 001
i.OO
• 999
.999
I.SS33
1.5359
I.S184
89
88
87
4
1
0 . 0698
0.0873
0 . 1047
.070
.087
.105
14-3
II. 5
9.57
.070
.087
.105
14.3
II-4
9.51
1.002
1.004
1.006
.998
.996
.995
1.5010
86
84
7
8
9
0.1222
0.1396
0.IS7I
.122
.139
.156
8.21
7.19
6.39
.123
.141
.158
8.14
7.12
6.31
1.008
1. 010
1. 012
.993
.990
.988
1.4486
I. 4312
1.4137
83
82
81
10
0.1745
.174
5. 76
.176
S.67
1. 015
.98s
1.3963
80
11
12
13
0.1920
0 . 2094
0.2269
.191
.208
.225
5. 24
4.81
4.45
.194
.213
.231
S.14
4.70
4.33
1. 019
1.022
1.026
.982
• 978
• 974
1.3788
1.3614
1.3439
79
78
77
14
IS
i6
0.2443
0.2618
0.2793
.242
.259
.276
3.63
■.HI
.287
4.01
3.73
3.49
1. 031
1.035
1.040
■.III
.961
1.3265
1.3090
I.291S
76
75
74
11
19
0.2967
0.3142
0.3316
.292
.309
.326
3.42
3.24
3.07
.306
.325
.344
3.27
3.08
2.90
1.046
1. 051
1.058
• 956
.946
I. 2741
1.2566
1.2392
73
72
71
20
U.3491
.342
2.92
.364
2.75
1.064
.940
1.2217
70
21
22
23
0.366s
0.3840
0.4014
.358
.375
.391
2.79
2.67
2.56
.384
.404
.424
2.61
2.48
2.36
1. 071
1.079
1.086
.934
.927
.921
I . 2043
I. 1868
I . 1694
69
68
67
t
24
0.4189
0.4363
0.4538
.407
Ml
2.46
2.37
2.28
1^1
.4S8
2.25
2.14
2.05
1.095
1. 103
1. 113
.914
.906
.899
1.1519
I. 1345
1.1170
66
64
11
29
0.4712
0.4887
0.S061
1P
.485
9.20
2.13
2.06
.510
.532
.554
1.96
1.88
1.80
1. 122
1 .133
1. 143
.891
.883
.875
I . 0996
1.0S21
1 . 0647
62
61
30
0.5236
.500
2.00
• 577
1.73
1. 155
.866
1.0472
60
31
32
33
0.S4"
0.558s
0.5760
.SIS
.530
.S4S
1.94
1.89
1.84
.601
.625
.649
1.66
1.60
1. 54
1.167
1. 179
1. 192
.857
.848
.839
1.0297
1.0123
0.9948
59
58
57
34
35
36
0.5934
0.6109
0.6283
.559
.574
.588
1.79
1. 74
1.70
.675
.700
.727
1.48
1-43
1.38
1.206
1.221
1.236
.829
.819
.809
0.9774
0.9599
0.942s
56
55
54
37
38
39
0.6458
0.6632
0.6807
.602
.616
.629
1.66
1.62
1.59
.754
.781
.810
1-33
1.28
1.23
1.252
1.269
1.287
.799
.788
.777
0.9250
0.9076
0.8901
53
52
SI
40
0.6981
.643
x.s6
.839
X.19
1.305
.766
0.8727
SO
41
42
43
0.7156
0.7330
0.7505
.656
.669
.682
1.52
1.49
1.47
.869
.900
• 933
1. 15
HI
1.07
1.32s
1.346
1.367
.7SS
■ 743
.731
0.8S52
0.8378
0.8203
49
48
47
44
45
0.7679
0.7854
.69s
.707
1.44
1. 41
.966
1. 00
1.04
1. 00
1.390
1.414
.719
.707
0.8029
0.7854
46
45
n cos
n sec
n cot
n tan
n CSC
n sin
Radians
Deg.
488 ELEMENTARY MATHEMATICAL ANALYSIS
Antilogakithms
0 ] I Ia|3<4l5|6|7l8 19
I 2 3 14 S 617 8 9l
•SO
•SI
•S2
■ S3
3162
3170
3177
3184
3192
3199
3206
3214
3221
3228
112
3 4 4
5 6 7
3236
33II
3388
3243
3319
3396
3251
3327
3404
3258
3334
3412
3266
3342
3420
3273
3350
3428
3281
3357
3436
3289
3365
3443
3296
3373
3451
3304
3381
3459
122
12 2
12 2
3 4 5
3 4 5
3 4 5
5 6 7
5 6 7
6 6 7
.S4
J
is7
.S8
■ S9
.6^
.6i
.62
.63
.64
it
3467
3548
3631
347S
3SS6
3639
3483
3565
3648
3491
3573
3656
3499
3581
3664
3S08
3589
3673
3516
3597
3681
3524
3606
3690
3532
3f4
3698
3540
3622
3707
12 2
12 2
r 2 3
3 4 5
3, 4 5
3 4 5
6 6 7
678
37IS
3802
3890
3724
3811
3899
3733
3819
3908
3828
3917
3750
3837
3926
3758
3846
3936
3767
38SS
3945
3776
3864
3954
3784
3873
3963
3793
3882
3972
I 2 3
I 2 3
123
3 4 5
4 4 5
4 5 5
i ' I
t ' 1
678
3981
3990
3999
4009
4018
4027
4036
4046
40SS
4064
I 2 3
4 5 6
678
4074
4169
4266
4083
4178
4276
4093
4188
4285
4102
4198
4295
4HI
4207
4305
4121
42J7
431s
4130
4227
4325
4140
4236
4335
4IS0
4246
4345
4159
4256
4355
1 2 3
I 2 3
I 2 3
456
456
4 5 6
7 8 9
7 8 9
7 8 9
436s
4467
4S7I
437S
4477
4581
438s
4487
4592
4395
4498
4603
4406
4508
4613
4416
4519
4624
4426
4529
4634
4436
4539
4645
4446
4550
4656
4457
4667
I 2 3
I 2 3
I 2 3
456
456
456
7 8 9
7 8 9
7 9 10
.69
4677
4786
4898
4688
4797
4909
4699
4808
4920
4710
4819
4932
4721
4831
4943
4732
4842
4955
4742
4853
4966
4'|3
4864
4977
4764
4875
4989
4775
4887
5000
I 2 3
I 2 3
I 2 3
4 5 7
467
5 6 7
8 9 10
8 9 lO
8 9 10
.70
5012
S023
S035
S047
S058
5070
S082
SO93
Sios
5117
I 2 4
S 6 7
8 9 II
.71
.72
.73
.74
S129
S248
S370
S140
5260
S383
5152
5272
5395
5164
S284
S408
5176
5297
5420
S188
S309
5433
5200
5321
S44S
5212
5333
5458
5224
5346
5470
5236
5358
S483
I 2 4
I 2 -4
I 3 4
5 6 7
5 6 7
568
8 10 II
9 10 II
9 10 II
549S
5623
S7S4
SS08
S636
S768
5S2I
5649
5781
5534
5662
5794
5546
5675
5808
SSS9
S689
5821
5572
5702
5834
5585
S7I5
5848
5598
5728
S86i
S6io
5741
587s
1 3 4
I 3 4
134
568
5 7 8
5 7 8
9 10 12
9 10 12
9 II 12
■77
.78
•79
5888
6026
6166
S902
6039
6180
5916
6053
6194
5929
6067
6209
5943
6081
6223
5957
6095
6237
5970
6109
6252
5984
6124
6266
5998
6138
6281
6012
6152
6295
I 3 4
I 3 4
I 3 4
5 7 8
678
679
10 II 12
10 II 13
10 II 13
.So
6310
6324
6339
6353
6368
6383
6397
6412
6427
6442
I 3 4
679
10 12 13
.81
.82
.83
6457
6607
6761
6471
6622
6776
6486
6637
6792
6501
6653
6808
6si6
6668
6823
6531
6683
6839
6855
6s6l
6714
6871
6577
6730
6887
6592
674s
6902
235
235
235
689
689
689
II 12 14
II 12 14
II 13 14
.85
.86
.i?
.88
.89
.90
6918
7079
7244
6934
7096
7261
6950
7112
7278
6966
7129
7295
6»82
7145
7311
6998
7161
7328
70IS
7178
7345
7031
7194
7362
7047
7211
7379
7063
7228
7396
235
235
235
6 8 10
7 8 10
7 8 10
11 13 15
12 13 15
12 13 15
7413
7S86
7762
7430
7603
7780
7447
7621
7798
7816
7482
7656
7834
7499
7674
7852
7516
7691
7870
7534
7709
7889
7SSI
7727
7907
7568
7745
7925
235
245
245
7 9 10
7 9 II
7 9 II
12 14 16
12 14 16
13 14 16
7943
7962
7980
7998
8017
803S
8054
8072
8091
8110
246
7 9 II
13 15 17
•91
.92
•93
8128
8318
8S1I
8147
8337
8S3I
8166
8356
8551
8185
8375
8570
8204
839s
8590
8222
8241
8433
8630
8260
8453
8650
8279
8472
8670
8299
8492
8690
246
246
246
8 9 II
8 10 12
8 10 12
13 IS 17
14 IS 17
14 16 18
■94
.96
.98
;99
8710
8913
9120
9333
9SS0
9772
8730
8933
9141
8750
8954
9162
8770
8974
91S3
8790
8995
9204
8810
9016
9226
8831
9036
9247
8851
9057
9268
8872
9078
9290
8892
909?
931 1
2 4 6
246
246
8 10 12
8 10 12
8 II 13
14 16 18
15 17 19
IS 17 19
9354
9572
9376
9594
^817
9397
9616
9840
9419
9638
986^
9661
0886
9462
9683
^08
9484
9705
mi
9506
9727
9954
9528
9750
997.7.
247
247
2 5 7
9 II 13
9 II 13
9 " 14
15 17 20
16 18 20
16 18 20
REVIEW OF SECONDARY SCHOOL ALGEBRA 489
Antilogamthms
10|ll2l3l4IS16|'7l8l9|l2 3l4 S 6' 7 89I
■00
01
■02
03
04
■U
■07
■08
•09
•10
■II
■12
13
■14
■It
■■\l
•19
■20
■21
■22
•23
•24
■25
■26
:S
■29
1000
1002
1005
1007
1009
1012
1014
1016
1019
IO21
0 0 I
I I I
222
1023
1047
1072
1026
lOSO
1074
102S
1052
1076
1030
1054
1079
1033
1057
1081
1035
1059
1084
1038
1062
1086
1040
1064
1089
1042
1067
IO91
1045
1069
1094
00 I
0 0 I
00 I
I I I
III
222
2 2 2
2 2 2
1096
1122
1 148
1099
II2S
IISI
1102
I127
1153
IlO/i
II30
II56
I107
1132
I159
1 109
II3S
1161
II12
1138
1 164
II14
I140
1167
III7
1 143
1 169
II19
I146
I172
0 1 I
oil
oil
112
112
112
2 2 2
2 2 2
2 2 2
1175
1202
1230
II78
1205
1233
1180
1208
1236
II83
I2II
1239
1 186
1213
1242
I189
1216
1245
I191
1219
1247
1194
1222
1250
II97
1225
1253
I199
1227
1256
0 I I
oil
oil
112
112
112
2 2 2
223
2 2 3
I2S9
1262
126s
1268
1271
1274
1276
1279
12S2
128s
0 1 I
112
223
1288
1318
1349
I29I
I32I
1352
1294
1324
1355
1297
1327
1358
1300
1330
1361
1303
1334
1365
1306
1337
1368
1309
1340
1371
13I2
1343
1374
1315
1346
1377
oil
oil
oil
12 2
12 2
12 2
223
223
233
1380
1413
1445
1384
I4I6
1449
1387
1419
1452
1390
1422
I4SS
1393
1426
1459
1396
1429
1462
1400
1432
1466
1403
1435
1469
1406
1439
1472
1409
1442
1476
0 1 I
0 I I
0 I I
12 2
12 2
12 2
233
233
233
1479
IS14
IS49
1483
ISI7
ISS2
i486
1521
1556
1489
1524
1560
1493
1528
1563
1496
1531
1567
1500
1535
1570
1503
1538
1574
1507
1542
1578
151O
IS4S
1581
oil
0 1 I
0 1 I
12 2
12 2
12 2
2 '3 3
233
3 3 3
ISSS
IS89
1592
1596
1600
1603
1607
1611
1614
1618
oil
12 2
3 3 3
1622
1660
1698
1626
1663
1702
1629
1667
1706
1633
167 1
1710
1637
1675
1714
1641
1679
1718
1644
1683
1722
1648
1687
1726
1652
1690
1730
1656
1694
1734
0 I I
0 I I
0 1 I
2 2 2
2 2 2
2 2 2
3 3 3
3 3^3
3 3 4
1738
1778
1820
1742
1782
1824
1746
1786
1828
1750
1791
1832
1754
1795
1837
1758
1799
1841
1762
1803
1845
1766
1807
1849
1770
1811
1854
1774
1816
1858
oil
oil
0 1 I
2 2 2
2 2 2
223
3 3 4
3 3 4
3 3 4
1862
190S
1950
1866
I9I0
1954
1871
1914
1959
1875
1919
1963
1879
1923
1968
1884
1928
1972
1888
1932
1977
1892
1936
1982
1897
1941
1986
1901
I94S
1991
0 I I
0 I I
223
223
2 2 3
3 3 4
3 4 4
3 4 4
.30
■31
•32
■33
•34
■35
■36
1995
2000
2004
2009
2014
2018
2023
2028
2032
2037
0 1 I
223
3 4 4
2042
2089
2138
2046
2094
2143
2051
2099
2148
2056
2104
2153
2061
2109
2158
2065
2113
2163
2070
2118
2168
2075
2123
2173
2080
2128
2178
2084
2133
2183
01 I
0 1 I
0 I I
223
223
2 2 3
3 4 4
3 4 4
3 4 4
21S8
2239
2291
2193
2244
2296
2198
2249
2301
2203
2254
2307
2208
2259
2312
2213
2265
2317
2218
2270
2323
2223
2275
2328
2228
2280
2333
2234
2286
2339
112
112
112
233
233
233
4 4 5
4 4 5
4 4 5
■11
■39
.40
2344
2399
2455
2350
2404
2460
2355
2410
2466
2360
2415
2472
2366
2421
2477
2371
2427
2483
2377
2432
2489
2382
2438
249s
2388
2443
2500
2393
2449
2506
112
112
l' I 2
233
233
233
4 4 5
4 4 5
4 5 5
2SI2
2518
2523
2529
2535
2541
2547
2553
2559
2564
112
234
4 5 5
•41
.42
.43
•44
it
■.ti
■49
2570
2630
2692
lilt
2698
2582
2642
2704
2588
2649
2710
2594
2655
2716
2600
2661
2723
2606
2667
2729
25l2
2673
2735
2618
2679
2742
2624
268s
2748
112
112
112
234
2 3 4
3 3 4
4 5 5
456
456
2754
2818
2884
2761
2825
2891
2767
2831
2897
2773
2838
2904
2780
2844
2911
2786
28S1
2917
2793
2858
2924
2799
2864
2931
2805
2871
2938
2812
2877
2944
112
I I 2
I I 2
3 3 4
3 3 4
3 3 4
4 5 6
5 5 S
5 5 6
29SI
3020
3090
2958
3027
3097
296s
3034
3105
2972
3041
3112
2979
3048
3119
2985
3055
3126
2992
3062
3133
2999
3069
3141
3006:3013
30763083
314813155
112
112
112
3 3 4
3 4 4
3 4 4
5 5 6
566
566
INDEX
(The numbers refer to the pages)
Abscissa, 33
Absolute value of complex num-
ber, 369
Addition formulas for sine and
cosine, 307-309
for tangent, 309
Additive properties of graphs,
142, 295-297
Aggregation, symbols of, 453
Algebraic scale, 3, 357
Alternating current curves, 384
et seq.
represented by complex
numbers, 384
Amplitude of complex number,
369
of S. H. M., 340
of sinusoid, 117
of uniform circular motion,
102
of wave, 345
Angle, 99
depression, 130
direction, 103
eccentric, 156
elevation, 130
epoch, 340, 345
initial side, 99
phase, 340, 345
that one line makes with an-
other, 313
vectorial, 103
Angles, congruent, 100
Angular magnitude, 99
units of measure, 100
velocity, 102
Anti-logarithm, 254
Approximation formulas, 209
Approximations, successive, 196
Argument of function, 12
of complex number, 369
Arithmetical mean, 213
progression, 213-216
triangle, 204
Asymptotes of hyperbola, 60,
165, 167
Auxiliary circles, 155
Axes of ellipse, 154
of hyperbola, 168
Binomial coefficients, graphical
representation of, 211,
212
theorem, 204 et seq.
Briggs, Henry, 236
system of logarithms, 245
Cartesian coordinates, 33
Cassinian ovals, 394
Catenary, 297
Change of base, 264, 265
of unit, 66, 70, foot note, 77
et seq., 285
Characteristic, 250, 251
Circle and circular functions.
Chap. IV, 97 et seq.
Circle, dipolar, 395
equation of, 97, 98
sine and cosine, 126-128
tangent to, 422, 428
through three points, 433
491
492
INDEX
(The numbers refer to the pages)
Circles, auxiliary, 155
Circular functions, 103 et seq.
graphical computation of,
106, 115
fundamental relations, 110,
304-318
law of, 132
motion, 102
Cologarithm, 254
Combinations, 199, 202, Chap.
VII
Common logarithms, 246
Complementary angles, 114, 118
Completing square, 463
Complex numbers, Chap. XII,
357 et seq.
defined, 363
laws of, 365
polar form, 369
typical form, 363
Composite angles, functions of,
310-312
Composition of two S. H. M.'s,
343
Compound harmonic motion, 334
interest, 220
law, 277
Computers rules, 328
Conditional equations, 138, 320-
327
Conies, 414, 417
con-focal, 441
sections. Chap. XIV, 399 et
seq.
Conjugate axis, 168
complex numbers, 367
hyperbolas, 170
Connecting rod motion, 355
Constants and variables, 15
Continuous function, 11
compounding of interest, 278
Coordinate paper, 27, 124, 271,
289
Coordinates, Chap. II, 23 et seq.
Cartesian, 34
orthogonal, 124
polar, 123, 434
rectangular, 33 et seq.
relation of polar and rectan-
gular, 136, 434
Cosecant, 103
Cosine, 103
curve, 117, 126
law, 321 ■
Cotangent, 103
Crest of sinusoid, 116
Cubical parabola, 52
Cubic equation, 192 et seq.
"Cut and Try," 149
Cycloid, 395
Damped vibrations, 299
Damping factor, 299
Decreasing function, 63
geometrical series, 221
DeMoivres theorem, 375
Descartes, Ren6, 34
Diameter of any curve, 440
of ellipse, 440
of parabola, 419
Direction angle, 103
Directrix of ellipse, 402, 415
of hyperbola, 408, 415
of parabola, 408, 413
Discontinuous function, 13, 27,
59
Distance of point from line,
426
Distributive law of multiplica-
tion, 205, 365
general, 456
INDEX
493
CThe numbers refer to the pages)
Double angle, functions of, 315
scale, 4-9, 21, 22, 266-276
of algebraic functions, 21
of logarithmic functions,
266-276
"e," 241, 245, 260, 277
Eccentric angle, 156
Eccentricity of earth's orbit, 403
of ellipse, 401
of hyperbola, 408
of parabola, 413, 415
Ellipse, 52 et seq., 399 et seq.,
Chaps. V and XIV.
axes of, 154
construction, 155, 158, 159
directrices, 402, 415
eccentricity, 402
focal radii, 399, 430
foci, 399
latus recturn, 404
parametric equation, 155
polar equation, 410
symmetrical equation of, 154.
tangent to, 429
vertices, 154
Ellipsograph, 158
Elliptic motion, 344, 388
Empirical curves, 46, 283, 291
formulas, 75
Envelope, 422
Epicycloid and epitrochoid, 397
Epoch angle, 340, 345, 348
Equations, conditional, 138
explicit, 154
quadratic, 462
systems, 186-192
simple, 461
single and simultaneous,
Chap. VI, 174
with given roots, 178
Even function, 119
Expansion, binomial, 205, 208
Exponential curves, 236-240,
260-264
equation, 240
function, Chap. IX, 234 et seq.
compared with power,
286-289
defined, 240, 243, 244
sums of, 295-299
Exponents, definition of, 466
irrational, 243, 244
laws of, 466
Factor theorem, 177
Factorial number, 200
Factoring, 454-460
fundamental theorem in, 459
Family of curves, 78
of lines, 421
Focal radii and foci, 393
of eUipse, 399, 430
of hyperbola, 406
radius of parabola, 414
Fractions, 460
Frequency of S. H. M., 340
of sinusoidal wave, 347
uniform circular motion, 102
Function, periodic, 26, 113
power, 48 et seq., 286
S. H. M., 339
trigononietric, 103
Functions, 10, 11
circular, Chap. IV, 97 et
seq., 103
continuous, 11
discontinuous, 13, 27, 59
even and odd, 119
explicit and implicit, 155, 174
exponential, 240,-243, 244,
286 et seq.
494
INDEX
(The numbers refer to the pages)
Functions, increasing and de-
creasing, 63, 147
General equation of second de-
gree, 440
Geometrical mean, 217
progression, 217 et seq.
Graphical computation, 16 et seq.
of circular functions, 106
of integral powers, 19
of logarithms, 237
of product, 16
of quotient, 17
of sq. roots, 18, 21
of squares, 18, 21
solution of cubic, 192
simultaneous equations,
183 et seq.
Graph of binomial coefficients,
211, 212
of complex number, 364
of cycloid, 396
of ellipse, 158, 159
of equation, 36
of functions of multiple an-
gles, 318, 319
of geometrical series, 236
of hyperbola, 167, 168
of hyperbolic functions, 297
of logarithmic and exponen-
tial curves, 236-240, 260
of parabolic arc, 420
of power function, 48-60, 64
of sinusoid, 115
of tangent and secant curves,
143-147
Graphs, suggestions on construc-
tion of, 27
nonnstatistical, 35
Half-angle, functions of, 315 .
Halley's law, 282
Haridonic analysis, 354
functions, 346
fundamental, 352
motion, Chap. XI, 339 et seq.
compound, 352
Hyperbola, Chap. V and XIV.
asymptotes, 165, 167
axes, 168
center, 168
conjugate, 170
construction of, 167
eccentricity, 408
foci and focal radii, 406
latus rectum, 408
parametric equations, 165,
167
polar equation, 410
rectangular, 58, 164
symmetrical equation, 166
vertices, 168
Hyperbolic curves, 52, 58
sine and cosine, 296
system of logarithms, 245
Hypocycloid and Hypo-trochoid,
397
i = V^^, 362
Identities, 110, 111, 138, 304r-317
Illustrations from science, 69-76
Image of curve, 57
Increasing function, 63, 147
progression, 214
Increment, logarithmic, 279
Infinite discontinuity, 59
geometrical progression, 221
Infinity, 69
Intercepts, 39, 40
Interest, compound, 220, 277
curve, 237
Interpolation, 252
Intersection of loci, 92, 182
INDEX
495
(The numbers refer to the pages)
Inverse of curve, 136
of straight line and circle, 136
trigonometric functions, 137,
360
Irrational numbers, 379
Lamellar motion, 88
Langley's law, 74
Latitude and longitude of a point,
33
Latus rectum of ellipse, 404
of hyperbola, 408
of parabola, 414
Law of circular functions, 132
of complex numbers, 365
of compound interest, 277
of exponential function, 288
of power function, 80-82
of sines, cosines, and tan-
gents, 320-327
Lead or lag, 349, 384
Legitimate transformations, 178
Lemniscate, 393
Limit, 221
Limiting lines of ellipse, 161
Loci, Chap. XIII, 387 et seq.
defined by focal radii, 393
Theorems on, 61, 62, 65, 85,
88, 135
Locus of points, 35, 36
of equation, 36
Logarithmic and exponential
functions, Chap. IX,
234 et seq.
coordinate paper, 289-295
curves, 236-240, 260-264
double scale, 266
functions, 240, 244
increment and decrement,
279-282, 299
tables, 252, 253
Logarithm of a number, 236, 244
Logarithms, common, 244
graph, 237-243
properties of, 247-250
systems of, 245
Mantissa, 250
Mean, arithmetical, 213
geometrical, 217
harmonical, 224
Modulus of complex number, 369
of decay, 281, 299
of logarithmic system, 264
Motion, circular, 102
compound harmonic, 352
connecting rod, 355
elliptic, 344, 388
shearing, 87
S. H. M., 339 et seq.
Naperian base, 245, 260, 277, 341
system of logs., 245
Napier, John, 234
Natural system of logarithms, 245
Negative angle, 100
functions of, 118, 119
Newton's law, 282
Node, 116
Normal, 136
equation of line, 136, 423
to ellipse, 430
to parabola, 420
Oblique triangles, 320-334
Odd functions, 119
Operators, 359
Ordinate of point, 33
Origin, 34
at vertex, 160, 415
Orthogonal systems, 124
Orthographic projection, 120-
123, 152, 265
496
INDEX
(The numbers refer to the pages)
Paper, logarithmic, 289 et seq.
polar, 124 et seq.
rectangular, 33 et seq.
semi-log, 271, 283 et seq.
Parabola, 52, 413
cubical, 52
polar equation, 414
properties of, 419
semi-cubical, 52
Parabolic curves, 49 et seq., 56, 289
Parameter, 155, 387
Parametric equations, 155
of cycloid, 396
of ellipse, 155
of hyperbola, 165, 166
Pascal's triangle, 204, 205
Periodic functions {see trig.-
fcns.), 26, 116
Period of S. H. M., 341
of simple pendulum, 342
of uniform circular mo-
tion, 102
of wave, 347
Permutations, 199-202
and combinations, Chap.
VII, 198 et seq.
Phase angle, 341, 348, 349
Plane triangles, 320-334
Polar coordinates, 123, 434
diagrams of periodic func-
tions, 126, 318
equation of ellipse, 410
of hyperbola, 410
of parabola, 414
of straight hne, 135
form of complex number, 369
relation to rectangular, 136,
434
Polynomial, 175
Positive and negative angle, 100,
119
Positive and negative coordi-
nates, 33
side of line, 427
Power function. Chap. Ill, 48 et
seq.
compared with exponen-
tial, 286-289
law of, 80-82
practical graph, 76
variation of, 62
Probability curve, 212
Products, special, 451, 452
Progressions, Chap. VIII, 213 et
seq.
arithmetical, 213-216
decreasing, 214
geometrical, 217-224
harmonical, 224, 225
Projection, orthographic, 120-
123, 152, 265
Proportionality factor, 68
Quadrants, 34
Quadratic equations, 462
systems of equations, 186
Questionable transformations,
178
Radian unit of measure, 101, 102
Radicals, reduction of, 471
Radius vector, 123
Ratio definition of conies, 414, 415
Rationalization, 472
Rational formulas, 75
numbers, 354
Rectangular coords, (see Coordi-
nates), Chap. II, 33
et seq.
Reflection of curve, 57
Remainder theorem, 175
Reversors, 361
Right angle system, 100
INDEX
497
(The numbers refer to the pages)
Root of any complex number, 376
of equation, 91
of function, 91, 177
of utoity, 377
Rotation of locus, 82, 133
polar coordinates, 133-135
rectangular, 434^436
of rigid body, 82
Scalar numbers, 358
Scale, 1, 3
algebraic, 3, 357
functions, 21
arithmetical, 3, 357
double, 4 et seq.
logarithmic, 266-276
uniform, 1
Tables, damped vibrations, 301,
302
logarithms, 252, 476
natural trig, functions, 107,
128, 129, 487
powers, 51
of "e," 263
Tangent, 103
graph, 143
law, 323
to circle, 422, 428
to curve, 260
to ellipse, 429
to parabola, 418
Theorems, binomial, 205 et seq.
factor, 176
functions of composite an-
gles, 310
on loci, 61, 65, 85, 135
remainder, 175
Transformations, legitimate and
questionable, 178
Translation, 82, 83
of any locus, 83, 85, 425
32
Translation of rigid body, 82
Transverse axis, 168
Triangle of reference, 103, 108
Triangles, solution of, 129, 320-
338
oblique, 320-338
right, 129-131
Trigonometric curves, 115, 117,
143-147, 319
functions, 103 et seq.
Trochoid, 397
Trochoidal waves, 349
Trough of sinusoid, 116
Uniform circular motion, 102
Unit, change of, 66, 70, 77 et
seq., 285
of angular measure, 101
Variables and constants, 15
and functions of variables.
Chap. I
Variation, 67
of power function, 62
Vector, 123
radius, 123
Vectorial angle, 103, 123
Velocity, angular, 102, 339
of wave, 348
Versors, 362
Vertices of ellipse, 154
of hyperbola, 168
Vibrations, damped, 299
Waves, Chap. XI, 339 et seq.
compound, 352
length of, 347
progressive, 344 et seq.
sinusoidal, 344 et seq.
stationary, 350
trochoidal, 349
Zero of function, 91
■i'li ! I! I!
iiiii:i:inii
liiiiiiii